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Series: Convergence and the Nonnegative Tests: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping

Example

The harmonic series is $\sum_{k \geq 1} 1 / k$ , the series from the starting index $1$ (Series, partial sums, convergence and the sum, divergence, and the tail series) of the family $a_{k} = 1 / ι (k)$ ; the index $0$ is excluded because $1 / 0$ has no value. It diverges, and its partial sums are unbounded above.

This is the case $p = 1$ of For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ . The two arguments below do not use that theorem: the first is the condensation argument, which is how For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ itself is proved and which here degenerates to something one can read off; the second is Oresme's block grouping from the fourteenth century, which uses no test at all and produces the explicit lower bound

$\sum_{k = 1}^{2^{n}} \frac{1}{k} \geq 1 + \frac{n}{2} (n \in N) .$

That bound is worth having on its own: it says the harmonic partial sums grow at least like a constant multiple of $n$ along the powers of $2$ , giving a concrete quantitative witness to their slow divergence.

Facts & Assumptions

Given: The family $a_{k} : = 1 / ι (k)$ for naturals $k \geq 1$ , with $ι (k)$ the canonical natural; its partial sums $A_{m} = \sum_{k = 1}^{m} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals are positive and order preserving: $0 < ι (j) \leq ι (k)$ for $1 \leq j \leq k$ ; and reciprocation reverses the order on the positives (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

Condensation: for a nonnegative nonincreasing family from $1$ , $\sum_{k \geq 1} x_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} x_{2^{j}}$ converges (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

A series whose terms do not tend to $0$ diverges (If a series converges then its terms tend to $0$ ).

[L4]

Powers of $2$ : $2^{j} \geq 1$ , $2^{j + 1} = 2 \cdot 2^{j}$ , and $2^{j} \cdot (1 / 2^{j}) = 1$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L5]

Splitting and monotonicity of finite sums, and the number of terms in $\sum_{k = p}^{q}$ , namely $q + 1 - p$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L6]

The principle of induction (The principle of mathematical induction); and for every real $x$ there is a natural $n$ with $ι (n) > x$ (Every complete ordered field is Archimedean).

[L7]

For a series of nonnegative terms: it converges if and only if the range of its partial sums is bounded above (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Lower bound, bounded below, bounded set).

Verification

technique · direct

1.1

Each $a_{k} = 1 / ι (k)$ is positive, and $a_{j} \geq a_{k}$ whenever $1 \leq j \leq k$ , since $0 < ι (j) \leq ι (k)$ and reciprocation reverses the order.

givenL1

1.2

For every $n \in N$ the block $\sum_{k = 2^{n} + 1}^{2^{n + 1}} 1 / k$ has $2^{n + 1} - 2^{n} = 2^{n}$ terms, each with index $k \leq 2^{n + 1}$ and hence each at least $1 / 2^{n + 1}$ ; so the block is at least $2^{n} \cdot \frac{1}{2^{n + 1}} = \frac{1}{2}$ .

L1L4L5

2.1

So the family is nonnegative and nonincreasing, and condensation applies to it.

step 1.1L2

2.2

An induction on $n$ gives $A_{2^{n}} \geq 1 + ι (n) / 2$ for every $n \in N$ . At $n = 0$ it reads $A_{1} = 1 \geq 1$ ; and if it holds at $n$ then, splitting at $2^{n}$ , $A_{2^{n + 1}} = A_{2^{n}} + \sum_{k = 2^{n} + 1}^{2^{n + 1}} 1 / k \geq 1 + ι (n) / 2 + 1 / 2 = 1 + ι (n + 1) / 2$ .

step 1.2L5L6

3.1

The condensed terms are $2^{j} a_{2^{j}} = 2^{j} \cdot \frac{1}{2^{j}} = 1$ for every $j \in N$ .

step 2.1L4

3.2

The range of the partial sums is not bounded above: given a real $M$ , choose a natural $n$ with $ι (n) > 2 M$ ; then $A_{2^{n}} \geq 1 + ι (n) / 2 > 1 + M > M$ .

step 2.2L6choose

4.1

The condensed series is therefore $\sum_{j \geq 0} 1$ , whose terms are constantly $1$ and so do not converge to $0$ ; it diverges.

step 3.1L3

5.1

By condensation, $\sum_{k \geq 1} 1 / k$ diverges. That is the first argument.

step 4.1step 2.1L2

6.1

Since the terms are nonnegative, the series diverges and its partial sums are unbounded above. That is the second argument, and it recovers the conclusion of step 5.1 without using any convergence test.

step 3.2step 1.1L7∎

Remarks

Why the two arguments are the same argument. Oresme's blocks are the blocks of the condensation proof, grouped from $2^{n} + 1$ to $2^{n + 1}$ , and the constant $1 / 2$ in step 1.2 is the constant that makes the condensed terms of step 3.1 equal to $1$ . The difference is bookkeeping: condensation states the grouping once and for all, for every nonincreasing family, and the block argument performs it for this one family.
The divergence is extremely slow, and the bound says how slow. To make the partial sum exceed $M$ the estimate of step 2.2 asks for about $2^{2 M}$ terms. That is why the harmonic series is the standard warning against reading convergence off numerical evidence.
The bound in step 2.2 is one sided. Nothing here says the partial sums are at most $1 + n / 2$ along powers of $2$ , and in fact they are not; the matching upper bound is the other half of the condensation estimate, and it is not needed for divergence.

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$\sum 1 / k^{2}$ converges with sum at most $2$ , by comparison with the telescoping $\sum 1 / (k (k - 1))$

Example

The series $\sum_{k \geq 1} 1 / k^{2}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and its sum is at most $2$ .

Convergence is the case $p = 2$ of For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ . What is added here is an elementary route that produces a numerical bound: for $k \geq 2$ ,

$\frac{1}{k^{2}} \leq \frac{1}{k (k - 1)} = \frac{1}{k - 1} - \frac{1}{k},$

so the terms from $k = 2$ on are dominated by a telescoping series of sum $1$ , and adding the first term $1$ gives the bound $2$ .

The bound is not the exact value. The sum is $π^{2} / 6$ , a fact requiring machinery this library develops much later; nothing below asserts or uses it.

Facts & Assumptions

Given: The families $c_{j} : = 1 / ι (j + 1)^{2}$ and $b_{j} : = 1 / ι (j + 1)$ for $j \in N$ , so that $\sum_{k \geq 1} 1 / k^{2}$ is the series of $(c_{j})$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^{m}$ , Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals are positive and order preserving, and reciprocation reverses the order on the positives (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

$\sum (b_{j} - b_{j + 1})$ converges whenever $(b_{j})$ converges, with sum $b_{0} - \lim_{j} b_{j}$ ( $\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$ ).

[L3]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Limits and Cauchy sequences of reals).

[L4]

Direct comparison (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ); and a series converges if and only if each of its tail series converges, the sum splitting as the initial partial sum plus the tail sum (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L5]

For a series of nonnegative terms the sum is the supremum of the partial sums, so every partial sum is at most the sum and the sum is at most any upper bound of the partial sums; and finite sums are monotone in their terms (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Lower bound, bounded below, bounded set, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

The series $\sum_{k \geq 1} 1 / k^{p}$ converges for rational $p > 1$ , so in particular at $p = 2$ , where $k^{2}$ is the integer power (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Rational powers $a^{r}$ of a positive base).

Verification

technique · direct

1.1

Every $c_{j}$ and every $b_{j}$ is positive.

givenL1

1.2

For every $j \in N$ : $b_{j} - b_{j + 1} = \frac{1}{ι (j + 1)} - \frac{1}{ι (j + 2)} = \frac{1}{ι (j + 1) ι (j + 2)}$ .

givenL1algebra

1.3

The sequence $(b_{j})$ converges to $0$ : given a rational $ε > 0$ , choose $n \geq 1$ with $1 / n < ε$ ; then $b_{j} = 1 / ι (j + 1) \leq 1 / n < ε$ for every $j$ with $j + 1 \geq n$ .

givenL1L3choose

2.1

Since $0 < ι (j + 1) \leq ι (j + 2)$ we have $ι (j + 1) ι (j + 2) \leq ι (j + 2)^{2}$ , hence $c_{j + 1} = \frac{1}{ι (j + 2)^{2}} \leq \frac{1}{ι (j + 1) ι (j + 2)} = b_{j} - b_{j + 1}$ for every $j$ .

step 1.2L1

2.2

By the telescoping lemma, $\sum_{j} (b_{j} - b_{j + 1})$ converges with sum $b_{0} - 0 = 1$ .

step 1.3L2

3.1

By comparison, $\sum_{j} c_{j + 1}$ converges, its terms being nonnegative and dominated by those of a convergent series.

step 2.1step 2.2step 1.1L4

3.2

Every partial sum of $\sum_{j} c_{j + 1}$ is at most the corresponding partial sum of $\sum_{j} (b_{j} - b_{j + 1})$ , which is at most $1$ ; so the sum of $\sum_{j} c_{j + 1}$ is at most $1$ .

step 2.1step 2.2L5

4.1

The series $\sum_{j} c_{j + 1}$ is the $1$ -st tail series of $\sum_{j} c_{j}$ , so $\sum_{j} c_{j}$ converges and its sum is $c_{0}$ plus that tail sum, that is at most $1 + 1 = 2$ .

step 3.1step 3.2L4

5.1

Since $\sum_{j} c_{j}$ is $\sum_{k \geq 1} 1 / k^{2}$ , that series converges with sum at most $2$ , in agreement with the case $p = 2$ of the $p$ -series theorem.

step 4.1givenL6∎

Remarks

The comparison starts at $k = 2$ and cannot start earlier. At $k = 1$ the dominating expression $1 / (k (k - 1))$ has a zero denominator, which is exactly why the argument is organised around the tail series and the first term is added back separately in step 4.1. That bookkeeping is where an off-by-one error would otherwise turn the bound $2$ into the false bound $1$ .
The telescoping comparison is sharper than it looks. The estimate $1 / k^{2} \leq 1 / (k (k - 1))$ loses only a factor $1 - 1 / k$ , so the bound $2$ sits not far above the true sum $π^{2} / 6$ . That comparison is orientation only; nothing on this page establishes the exact value, and nothing on this page uses it.

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Geometric sums computed: $\sum_{k \geq 1} 2^{- k} = 1$ and $\sum_{k \geq 0} (- 1 / 3)^{k} = 3 / 4$

Example

Two geometric sums, computed from For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges and stated with the starting index made explicit:

$\sum_{k \geq 1} 2^{- k} = 1, \sum_{k \geq 0} (- \frac{1}{3})^{k} = \frac{3}{4} .$

Both series converge, so both symbols denote (Series, partial sums, convergence and the sum, divergence, and the tail series).

The first is the one that is easy to get wrong. The theorem gives $\sum_{k \geq 0} 2^{- k} = 1 / (1 - 1 / 2) = 2$ , a series whose first term is $2^{0} = 1$ . The series above starts at $k = 1$ and therefore omits that term, so its sum is $2 - 1 = 1$ , not $2$ . A geometric series is not determined by its ratio alone; the starting index has to be said, and here it is.

Facts & Assumptions

Given: The real numbers $1 / 2$ and $- 1 / 3$ , and the integer powers $r^{k}$ (Integer powers $a^{m}$ ).

[L1]

For $∣ r ∣ < 1$ the series $\sum r^{k}$ from the starting index $0$ converges with sum $1 / (1 - r)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Absolute value: $∣ 1 / 2 ∣ = 1 / 2$ and $∣ - 1 / 3 ∣ = 1 / 3$ , both less than $1$ (Basic properties of the absolute value).

[L3]

A series converges if and only if each of its tail series converges, and then the sum splits as $s_{N}$ plus the sum of the $N$ -th tail series (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L4]

$r^{0} = 1$ for every real $r$ (Integer powers $a^{m}$ , Laws of integer exponents).

Verification

technique · direct

1.1

Since $∣ 1 / 2 ∣ = 1 / 2 < 1$ , the series $\sum_{k \geq 0} (1 / 2)^{k}$ converges with sum $1 / (1 - 1 / 2) = 2$ .

givenL1L2algebra

1.2

Since $∣ - 1 / 3 ∣ = 1 / 3 < 1$ , the series $\sum_{k \geq 0} (- 1 / 3)^{k}$ converges with sum $1 / (1 - (- 1 / 3)) = 1 / (4 / 3) = 3 / 4$ , which is the second claim.

givenL1L2algebra

1.3

The series $\sum_{k \geq 1} 2^{- k}$ is the $1$ -st tail series of $\sum_{k \geq 0} (1 / 2)^{k}$ , its terms being $(1 / 2)^{j + 1} = 2^{- (j + 1)}$ for $j \in N$ .

givenL3L4

2.1

The first partial sum of $\sum_{k \geq 0} (1 / 2)^{k}$ is $s_{1} = (1 / 2)^{0} = 1$ , so by the splitting identity the tail sum is $2 - 1 = 1$ , which is the first claim.

step 1.1step 1.3L3L4∎

Remarks

The two computations use the theorem in different regimes of sign. The first has a positive ratio and a monotone sequence of partial sums; the second has a negative ratio, so its partial sums oscillate around the limit rather than climbing to it. The theorem covers both without a case split, because its hypothesis is on $∣ r ∣$ and its proof runs through $s_{n} = (1 - r^{n}) / (1 - r)$ , which is indifferent to the sign of $r$ .
Where the starting index bites. Every application of a geometric comparison on this page and its companion begins by fixing which index the comparison series starts at, precisely because the sum changes by the omitted terms while the fact of convergence does not.

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$\sum_{k \geq 1} 1 / (k (k + 1)) = 1$

Example

$\sum_{k \geq 1} \frac{1}{k (k + 1)} = 1.$

The series converges, so the symbol denotes (Series, partial sums, convergence and the sum, divergence, and the tail series), and its sum is exactly $1$ . The reason is the partial fraction identity

$\frac{1}{k (k + 1)} = \frac{1}{k} - \frac{1}{k + 1},$

which makes the series telescoping with $b_{k} = 1 / k$ : the partial sums are $1 - 1 / (n + 1)$ , and $1 / (n + 1) \to 0$ .

Compare $\sum_{k \geq 1} 1 / k$ , which diverges (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping). The single extra factor $k + 1$ in the denominator is what separates the two.

Facts & Assumptions

Given: The sequence $b_{j} : = 1 / ι (j + 1)$ for $j \in N$ , so that $b_{j} = 1 / k$ at $k = j + 1$ ; and the family $d_{k} : = 1 / (k (k + 1))$ for naturals $k \geq 1$ , so that $\sum_{k \geq 1} d_{k}$ is the series of $j \mapsto d_{j + 1}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals $ι (j + 1)$ are positive, and reciprocals of positives are positive (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

$\sum (b_{j} - b_{j + 1})$ converges if and only if $(b_{j})$ converges, and then its sum is $b_{0} - \lim_{j} b_{j}$ ( $\sum (b_{k} - b_{k + 1})$ converges iff $(b_{k})$ converges, with sum $b_{0} - \lim b_{k}$ ).

[L3]

Verification

technique · direct

1.1

For every $j \in N$ : $b_{j} - b_{j + 1} = \frac{1}{ι (j + 1)} - \frac{1}{ι (j + 2)} = \frac{ι (j + 2) - ι (j + 1)}{ι (j + 1) ι (j + 2)} = \frac{1}{ι (j + 1) ι (j + 2)}$ .

givenL1algebra

1.2

The sequence $(b_{j})$ converges to $0$ : given a rational $ε > 0$ , choose $n \geq 1$ with $1 / n < ε$ ; then for every $j$ with $j + 1 \geq n$ we have $0 < b_{j} \leq 1 / n < ε$ .

givenL1L3choose

2.1

The term of $\sum_{k \geq 1} d_{k}$ at index $j$ is $d_{j + 1} = \frac{1}{ι (j + 1) ι (j + 2)}$ , so it equals $b_{j} - b_{j + 1}$ ; the two series are the same series.

step 1.1given

2.2

By the telescoping lemma, $\sum_{j} (b_{j} - b_{j + 1})$ converges with sum $b_{0} - 0 = 1 / ι (1) = 1$ .

step 1.2L2L1

3.1

Therefore $\sum_{k \geq 1} 1 / (k (k + 1))$ converges with sum $1$ .

step 2.1step 2.2∎

Remarks

The value $1$ comes from the first term of $(b_{j})$ , not from the first term of the series. The telescoping lemma gives $b_{0} - \lim_{j} b_{j}$ , and here $b_{0} = 1 / 1 = 1$ while the first term of the series is $d_{1} = 1 / 2$ . Reading the sum off the wrong one of those two numbers is the standard error, and it is why the lemma states the value in terms of $b_{0}$ explicitly.
Every telescoping identity is an identity between finite sums. Nothing about limits enters step 1.1; the only limit in the argument is $1 / (n + 1) \to 0$ , which is the Archimedean property. That is the general shape of every telescoping computation on this page.

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Condensation reduces $\sum 1 / k^{p}$ to a geometric series with ratio $2^{1 - p}$

Example

Let $p \in Q$ with $p > 0$ . Condensation (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges) applied to the family $a_{k} = 1 / k^{p}$ , $k \geq 1$ , produces a geometric series of ratio $2^{1 - p}$ :

$2^{j} a_{2^{j}} = 2^{j} (2^{j})^{- p} = 2^{(1 - p) j} = (2^{1 - p})^{j} (j \in N) .$

So the whole $p$ -series family collapses onto the single question of when a geometric ratio is below $1$ , and the threshold $p = 1$ is where $2^{1 - p} = 2^{0} = 1$ . That is the computation behind For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , displayed here on its own and instantiated at three exponents:

$p$	ratio $2^{1 - p}$	condensed series	verdict
$1 / 2$	$2^{1 / 2}$	diverges, ratio $> 1$	$\sum_{k \geq 1} k^{- 1 / 2}$ diverges
$1$	$1$	diverges, terms constantly $1$	$\sum_{k \geq 1} 1 / k$ diverges
$2$	$1 / 2$	converges, sum $2$	$\sum_{k \geq 1} 1 / k^{2}$ converges

Facts & Assumptions

Given: A rational $p > 0$ and the family $a_{k} : = ι (k)^{- p}$ for naturals $k \geq 1$ (Rational powers $a^{r}$ of a positive base, Canonical naturals are positive and strictly increasing).

[L1]

[L2]

Rational powers of a positive base: $a^{r + s} = a^{r} a^{s}$ , $(a^{r})^{s} = a^{r s}$ , $a^{- r} = 1 / a^{r}$ , $a^{r} > 0$ ; the integer power agrees with the rational power at an integer exponent, since $a^{1 / 1} = a$ ; and $a^{0} = 1$ (Laws of rational exponents, Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Integer powers $a^{m}$ ).

[L3]

Monotonicity of rational powers: for $a > 1$ and rationals $r < s$ , $a^{r} < a^{s}$ ; and for rational $t > 0$ , $0 < a < b$ implies $a^{t} < b^{t}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L4]

The geometric series $\sum_{j \geq 0} r^{j}$ converges exactly when $∣ r ∣ < 1$ , with sum $1 / (1 - r)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L5]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ); the canonical naturals are positive and order preserving, and reciprocation reverses the order on the positives (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

Verification

technique · direct

1.1

Each $a_{k} = ι (k)^{- p}$ is positive, and $a_{j} \geq a_{k}$ whenever $1 \leq j \leq k$ , since $ι (j)^{p} \leq ι (k)^{p}$ for $p > 0$ and reciprocation reverses the order; so condensation applies.

givenL3L5L1

1.2

For every $j \in N$ : $2^{j} a_{2^{j}} = 2^{j} (2^{j})^{- p} = 2^{j} \cdot 2^{- j p} = 2^{j - j p} = 2^{(1 - p) j} = (2^{1 - p})^{j}$ , reading each integer exponent as a rational one.

L2algebra

2.1

So the condensed series is the geometric series of ratio $r : = 2^{1 - p}$ , which is positive; and $r < 1$ exactly when $1 - p < 0$ , since $2 > 1$ makes $t \mapsto 2^{t}$ strictly increasing and $2^{0} = 1$ .

step 1.2L2L3

3.1

At $p = 2$ : the ratio is $2^{- 1} = 1 / 2$ , so the condensed series converges with sum $1 / (1 - 1 / 2) = 2$ , and $\sum_{k \geq 1} 1 / k^{2}$ converges.

step 1.2step 2.1L1L4L5

3.2

At $p = 1$ : the ratio is $2^{0} = 1$ , the condensed terms are constantly $1$ , so the condensed series diverges and $\sum_{k \geq 1} 1 / k$ diverges.

step 1.2step 2.1L1L4L5

3.3

At $p = 1 / 2$ : the ratio is $2^{1 / 2}$ , which exceeds $1$ because $2 > 1$ and $1 / 2 > 0$ ; so the condensed series diverges and $\sum_{k \geq 1} k^{- 1 / 2}$ diverges.

step 1.2step 2.1L1L3L4L5

4.1

The three verdicts agree with the $p$ -series theorem, whose content is exactly step 2.1 together with the geometric threshold.

step 3.1step 3.2step 3.3L5∎

Remarks

The sum of the condensed series is not the sum of the original. At $p = 2$ the condensed series sums to $2$ while $\sum_{k \geq 1} 1 / k^{2}$ sums to $π^{2} / 6$ . Condensation preserves the fact of convergence and nothing numerical, which is visible in its proof: the two estimates there differ by a factor $2$ .
Why the exponent has to be rational. The identity in step 1.2 is a chain of rational-exponent laws, and $2^{1 - p}$ is meaningful here only because $1 - p$ is rational (Rational powers $a^{r}$ of a positive base). The same computation with a real exponent is the standard one, and it waits for the exponential function.

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Abel-Dini applied to $\sum 1 / k$ : $\sum 1 / (k s_{k})$ still diverges while $\sum 1 / (k s_{k}^{2})$ converges

Example

Take $a_{k} : = 1 / ι (k + 1)$ for $k \in N$ , so that $\sum a_{k}$ is the harmonic series $\sum_{k \geq 1} 1 / k$ , which has positive terms and diverges (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping). Its inclusive partial sums are the harmonic numbers

$S_{n} = \sum_{k = 0}^{n} a_{k} = \sum_{k = 1}^{n + 1} \frac{1}{k} = H_{n + 1} (n \in N),$

all of them positive. The Abel-Dini theorem (For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges) then says that

$\sum_{n} \frac{a_{n}}{S_{n}} = \sum_{n} \frac{1}{(n + 1) H_{n + 1}} diverges, \sum_{n} \frac{a_{n}}{S_{n}^{2}} = \sum_{n} \frac{1}{(n + 1) H_{n + 1}^{2}} converges .$

Classically these are written $\sum_{k \geq 1} 1 / (k H_{k})$ and $\sum_{k \geq 1} 1 / (k H_{k}^{2})$ , with $H_{k} = 1 + 1 / 2 + \dots + 1 / k$ .

What the pair shows. The harmonic series is a familiar slowly divergent explicit series, and dividing its terms by the running total produces something that diverges more slowly still. Dividing by the square of the running total overshoots into convergence. So exponent $1$ gives a divergent member and exponent $2$ a convergent one. The absence of a slowest divergent positive series comes from applying Abel-Dini again to the newly produced divergent series, not from a last-exponent claim about this fixed pair.

Facts & Assumptions

Given: The sequence $a_{k} : = 1 / ι (k + 1)$ , $k \in N$ , and its inclusive partial sums $S_{n} = \sum_{k = 0}^{n} a_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals are positive, so each $a_{k}$ is positive and each $S_{n}$ is a sum of positive terms (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

The harmonic series $\sum_{k \geq 1} 1 / k$ diverges (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping), and it is by definition the series of $j \mapsto 1 / ι (j + 1)$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L3]

Abel-Dini: for a sequence of positive terms whose series diverges, with $S_{n}$ the inclusive partial sums, $\sum_{n} a_{n} / S_{n}$ diverges and $\sum_{n} a_{n} / S_{n}^{2}$ converges (For a divergent series of positive terms with partial sums $s_{k}$ , the series $\sum a_{k} / s_{k}$ diverges and $\sum a_{k} / s_{k}^{2}$ converges, Integer powers $a^{m}$ ).

Verification

technique · direct

1.1

Every term $a_{k} = 1 / ι (k + 1)$ is positive.

givenL1

1.2

The series $\sum a_{k}$ is the harmonic series $\sum_{k \geq 1} 1 / k$ and therefore diverges.

givenL2

1.3

Its inclusive partial sums are $S_{n} = \sum_{k = 0}^{n} 1 / ι (k + 1) = \sum_{k = 1}^{n + 1} 1 / k = H_{n + 1}$ , a reindexing of the sum by $k \mapsto k + 1$ .

givenL1

2.1

The hypotheses of Abel-Dini are met by $(a_{k})$ : positive terms and a divergent series.

step 1.1step 1.2L3

3.1

Therefore $\sum_{n} \frac{a_{n}}{S_{n}} = \sum_{n} \frac{1}{ι (n + 1) H_{n + 1}}$ diverges.

step 2.1step 1.3L3

4.1

And $\sum_{n} \frac{a_{n}}{S_{n}^{2}} = \sum_{n} \frac{1}{ι (n + 1) H_{n + 1}^{2}}$ converges.

step 2.1step 1.3L3∎

Remarks

This is the concrete form of the no-slowest-series obstruction. The general statement is that no divergent series of positive terms is eventually dominated by every other; here it is exhibited for the standard candidate. Anyone proposing the harmonic series as a universal comparison series is answered by the first of the two conclusions.
No growth estimate for $H_{k}$ is used or needed. The classical statement $H_{k} \approx \log k$ would make both conclusions look like instances of the $p$ -series with a logarithmic correction, but neither the logarithm nor that estimate is available in this library at this point, and the theorem does not require them: it needs only that the running totals are positive, nondecreasing and unbounded.

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A series with ratio limit exactly $1$ that Raabe decides

Example

Take $a_{k} : = 1 / ι (k + 1)^{2}$ for $k \in N$ , so that $\sum a_{k}$ is $\sum_{k \geq 1} 1 / k^{2}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

its ratios $q_{k} = a_{k + 1} / a_{k} = \frac{(k + 1)^{2}}{(k + 2)^{2}}$ converge to $1$ , so neither half of the ratio test applies (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence);
its Raabe expression is exactly $R_{k} = (k + 1) (\frac{a_{k}}{a_{k + 1}} - 1) = \frac{(k + 2)^{2} - (k + 1)^{2}}{k + 1} = \frac{2 k + 3}{k + 1} = 2 + \frac{1}{k + 1},$ which exceeds $2$ at every index, so ${lim inf}_{k} R_{k} \geq 2 > 1$ and Raabe's test gives convergence (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence).

This is the smallest honest illustration that Raabe's test decides series the ratio test cannot. The verdict agrees with For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ at $p = 2$ , as it must.

Facts & Assumptions

Given: The sequence $a_{k} : = 1 / ι (k + 1)^{2}$ , $k \in N$ ; its ratios $q_{k} = a_{k + 1} / a_{k}$ ; and its Raabe expression $R_{k} = (k + 1) (a_{k} / a_{k + 1} - 1)$ (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence, Integer powers $a^{m}$ , Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals are positive, so every $a_{k}$ is positive; reciprocation on the positives is order reversing; and $x^{2} = x \cdot x$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L2]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ , so $1 / ι (k + 1) \to 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Limits and Cauchy sequences of reals).

[L3]

Algebra of limits: sums, products and quotients of convergent sequences converge, the quotient requiring a nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients).

[L4]

The ratio test: its convergence half needs ${lim sup}_{k} q_{k} < 1$ and its divergence half needs ${lim inf}_{k} q_{k} > 1$ (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

[L5]

Raabe's test: ${lim inf}_{k} R_{k} > 1$ gives convergence (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence).

[L6]

Limit superior and inferior in $\overline{R}$ , their existence for every sequence, and the descriptions ${lim sup}_{k} x_{k} = \inf {s_{n}}$ , ${lim inf}_{k} x_{k} = \sup {i_{n}}$ of the tail bounds (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L7]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Verification

technique · direct

1.1

Every $a_{k} = 1 / ι (k + 1)^{2}$ is positive, so the ratios and the Raabe expression are defined.

givenL1

2.1

The ratios are $q_{k} = \frac{1 / ι (k + 2)^{2}}{1 / ι (k + 1)^{2}} = \frac{ι (k + 1)^{2}}{ι (k + 2)^{2}} = {(1 - \frac{1}{ι (k + 2)})}^{2}$ .

step 1.1L1algebra

2.2

The Raabe expression is $R_{k} = (k + 1) (\frac{ι (k + 2)^{2}}{ι (k + 1)^{2}} - 1) = \frac{ι (k + 2)^{2} - ι (k + 1)^{2}}{ι (k + 1)} = \frac{2 ι (k) + 3}{ι (k) + 1} = 2 + \frac{1}{ι (k + 1)}$ .

step 1.1L1algebra

3.1

Since $1 / ι (k + 2) \to 0$ , the product rule gives $q_{k} \to (1 - 0)^{2} = 1$ .

step 2.1L2L3

3.2

From step 2.2, $R_{k} > 2$ for every $k \in N$ , the added term $1 / ι (k + 1)$ being positive.

step 2.2L1

4.1

The convergence half of the ratio test does not apply: if ${lim sup}_{k} q_{k} < 1$ , then with $t$ real and ${lim sup}_{k} q_{k} < t < 1$ some tail supremum would be below $t$ , putting $q_{k} \leq t < 1$ for all large $k$ and contradicting $q_{k} \to 1$ .

step 3.1L4L6

4.2

The divergence half does not apply either: if ${lim inf}_{k} q_{k} > 1$ , some tail infimum would exceed $1$ , putting $q_{k} \geq c > 1$ for all large $k$ and again contradicting $q_{k} \to 1$ .

step 3.1L4L6

4.3

On the other hand $2$ is a lower bound of ${R_{k} : k \geq 0}$ , so the tail infimum $i_{0} \geq 2$ and ${lim inf}_{k} R_{k} \geq 2 > 1$ .

step 3.2L6

5.1

Raabe's test therefore gives convergence of $\sum a_{k}$ , that is of $\sum_{k \geq 1} 1 / k^{2}$ , in agreement with the case $p = 2$ of the $p$ -series theorem.

step 4.3step 1.1L5L7∎

Remarks

The Raabe expression here is exact, not asymptotic. Step 2.2 computes $R_{k} = 2 + 1 / (k + 1)$ on the nose, so no limit is needed to apply the test: a single inequality $R_{k} > 2$ at every index already forces ${lim inf}_{k} R_{k} \geq 2$ . That is why this witness is the cleanest available one.
Why the ratio test must fail here. The ratios of any $p$ -series tend to $1$ whatever $p$ is, so a criterion reading only $lim sup$ and $lim inf$ of the ratios cannot separate the convergent $p$ -series from the divergent ones. Raabe reads the rate at which the ratios approach $1$ , which is exactly the missing information, and that rate is $2 / k$ up to smaller terms when $p = 2$ .

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

$a_{k} = 2^{- k + (- 1)^{k}}$ has ratio limsup $2$ and liminf $1 / 8$ , so the ratio test fails, while the root test gives convergence

Statement refuted

Refuted claim: whenever the root test decides a series, the ratio test decides it too; equivalently, the ratio test is no weaker than the root test.

The claim is refuted by the sequence usually written $a_{k} = 2^{- k + (- 1)^{k}}$ . Precisely, let $(s_{k})$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , let $t_{k} : = 2$ when $s_{k} = 1$ and $t_{k} : = 1 / 2$ when $s_{k} = - 1$ , and put

$a_{k} : = 2^{- k} t_{k} (k \in N) .$

Its ratio and root families, in the shifted form used throughout, $q_{k} = ∣ a_{k + 1} / a_{k} ∣$ and $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , satisfy

${lim inf}_{k} q_{k} = \frac{1}{8}, {lim sup}_{k} q_{k} = 2, {lim sup}_{k} ρ_{k} = \frac{1}{2},$

as computed in FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence. So the root test gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ , while neither half of the ratio test applies: its convergence half needs ${lim sup}_{k} q_{k} < 1$ and $2$ is not below $1$ , and its divergence half needs ${lim inf}_{k} q_{k} > 1$ and $1 / 8$ is not above $1$ .

This is the concrete form of the strict dominance recorded in Whenever the ratio test decides, the root test decides the same way, and the converse fails.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ; $t_{k} : = 2$ when $s_{k} = 1$ and $t_{k} : = 1 / 2$ when $s_{k} = - 1$ ; $a_{k} : = 2^{- k} t_{k}$ ; and the families $q_{k} = ∣ a_{k + 1} / a_{k} ∣$ , $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , Rational powers $a^{r}$ of a positive base).

[L1]

The alternating sequence satisfies $∣ s_{k} ∣ = 1$ for every $k$ , so each $s_{k}$ is $1$ or $- 1$ and $t_{k}$ is well defined with $t_{k} > 0$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , Basic properties of the absolute value).

[L2]

$2^{- k} > 0$ for every $k \in N$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Laws of integer exponents).

[L3]

For this sequence, ${lim inf}_{k} q_{k} = 1 / 8$ , ${lim sup}_{k} q_{k} = 2$ and the root family converges to $1 / 2$ , so ${lim sup}_{k} ρ_{k} = 1 / 2$ (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence).

[L4]

The root test: ${lim sup}_{k} ρ_{k} < 1$ gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L5]

The ratio test: its convergence half needs ${lim sup}_{k} q_{k} < 1$ and its divergence half needs ${lim inf}_{k} q_{k} > 1$ ; those are its only two criteria (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

[L6]

Limit superior and inferior exist in $\overline{R}$ for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ); a series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series).

Counterexample

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ , so $t_{k}$ is defined and positive, and $a_{k} = 2^{- k} t_{k} > 0$ ; in particular $a_{k} \neq 0$ , so both the ratio and the root families are defined and $∣ a_{k} ∣ = a_{k}$ .

givenL1L2

2.1

For this sequence ${lim sup}_{k} q_{k} = 2$ and ${lim inf}_{k} q_{k} = 1 / 8$ .

step 1.1L3

2.2

For this sequence ${lim sup}_{k} ρ_{k} = 1 / 2$ .

step 1.1L3

3.1

Since $1 / 2 < 1$ , the root test applies and gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ , hence of $\sum_{k \geq 1} a_{k}$ , the terms being positive.

step 2.2step 1.1L4L6

3.2

The convergence half of the ratio test does not apply, since ${lim sup}_{k} q_{k} = 2$ and $2 < 1$ is false.

step 2.1L5

3.3

The divergence half does not apply either, since ${lim inf}_{k} q_{k} = 1 / 8$ and $1 / 8 > 1$ is false.

step 2.1L5

4.1

So the root test decides this series and the ratio test decides nothing about it, refuting the claim.

step 3.1step 3.2step 3.3∎

Remarks

The two families are computed once, on the previous page, and cited here. The four limit quantities for this sequence are established in the refutation of FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence, where the same witness shows that the outer inequalities of the ratio-to-root chain are strict. Nothing is recomputed here; what is added is the reading of those numbers through the two tests.
Why the roots behave and the ratios do not. The exponent of $a_{k}$ is $- k + (- 1)^{k}$ . Taking an $(k + 1)$ -st root divides that exponent by $k + 1$ , so the bounded oscillation contributes $2^{\pm 1 / (k + 1)}$ , which tends to $1$ ; forming a ratio differences the exponent, and a bounded oscillation does not shrink under differencing but doubles.
This does not make the root test universal. The companion counterexample with root limit exactly $1$ shows the root test has its own blind spot, and FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test shows no test on this page can avoid having one.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

$\sum k^{- 1 / 2}$ diverges and $\sum k^{- 2}$ converges, and both have root limit exactly $1$

Statement refuted

Refuted claim: the value ${lim sup}_{k} ∣ a_{k + 1} ∣^{1 / (k + 1)} = 1$ determines the behaviour of $\sum_{k \geq 1} a_{k}$ ; that is, any two series with root quantity equal to $1$ either both converge or both diverge.

The claim is refuted by the two families

$a_{k} : = k^{- 1 / 2}, b_{k} : = k^{- 2} (k \geq 1),$

rational powers of the canonical naturals (Rational powers $a^{r}$ of a positive base). Both have root quantity exactly $1$ , while $\sum_{k \geq 1} a_{k}$ diverges and $\sum_{k \geq 1} b_{k}$ converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , at $p = 1 / 2$ and $p = 2$ ).

So the third clause of Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing is not a gap in the proof: at $lim sup = 1$ nothing whatever follows, and the two witnesses here are on opposite sides.

Facts & Assumptions

Given: The families $a_{k} : = ι (k)^{- 1 / 2}$ and $b_{k} : = ι (k)^{- 2}$ for naturals $k \geq 1$ ; the sequence $u_{j} : = ι (j + 1)^{1 / (j + 1)}$ , $j \in N$ ; and the root families $α_{j} : = a_{j + 1}^{1 / (j + 1)}$ , $β_{j} : = b_{j + 1}^{1 / (j + 1)}$ (Rational powers $a^{r}$ of a positive base, Canonical naturals are positive and strictly increasing).

[L1]

$1 \leq n^{1 / n}$ for every natural $n \geq 1$ , and $u_{j} = (j + 1)^{1 / (j + 1)} \to 1$ ( $n^{1 / n} \to 1$ ).

[L2]

Laws of rational exponents on a positive base: $(x^{r})^{s} = x^{r s}$ , $x^{- r} = 1 / x^{r}$ , $x^{r} > 0$ , and $1^{r} = 1$ (Laws of rational exponents, Rational powers $a^{r}$ of a positive base).

[L3]

Monotonicity of rational powers: for rational $t > 0$ , $0 < x \leq y$ implies $x^{t} \leq y^{t}$ ; and for $x \geq 1$ and rationals $r < s$ , $x^{r} \leq x^{s}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L4]

The squeeze theorem, and the product and quotient rules for limits, the quotient requiring a nonzero limit and nonzero denominators (The squeeze theorem, Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L5]

A sequence converging to a real $c$ has $lim sup = lim inf = c$ (A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , and diverges to $\pm \infty$ iff both equal $\pm \infty$ , Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[L6]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

The canonical naturals are positive with $ι (k) \geq 1$ for $k \geq 1$ ; reciprocation reverses the order on the positives; and $∣ x ∣ = x$ for $x \geq 0$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value).

Counterexample

technique · direct

1.1

For $k \geq 1$ we have $ι (k) \geq 1 > 0$ , so $a_{k}$ and $b_{k}$ are positive and equal to their own absolute values; and $u_{j} \geq 1 > 0$ for every $j$ .

givenL1L2L7

1.2

The root family of $(b_{k})$ is $β_{j} = (ι (j + 1)^{- 2})^{1 / (j + 1)} = (ι (j + 1)^{1 / (j + 1)})^{- 2} = 1 / u_{j}^{2}$ .

givenL2

1.3

The series $\sum_{k \geq 1} k^{- 1 / 2}$ is the $p$ -series at $p = 1 / 2$ , and $1 / 2 > 1$ is false, so it diverges.

givenL6

1.4

The series $\sum_{k \geq 1} k^{- 2}$ is the $p$ -series at $p = 2$ , and $2 > 1$ , so it converges.

givenL6

2.1

Since $ι (j + 1) \geq 1$ and $- 1 < - 1 / 2 < 0$ , we have $ι (j + 1)^{- 1} \leq ι (j + 1)^{- 1 / 2} \leq ι (j + 1)^{0} = 1$ .

step 1.1L3L2

2.2

Since $u_{j} \to 1$ , the product rule gives $u_{j}^{2} \to 1$ , and the quotient rule then gives $β_{j} \to 1$ ; so ${lim sup}_{j} β_{j} = 1$ .

step 1.2L1L4L5

3.1

The root family of $(a_{k})$ is $α_{j} = (ι (j + 1)^{- 1 / 2})^{1 / (j + 1)} = ι (j + 1)^{- 1 / (2 (j + 1))}$ , and applying the same exponent to the two bounds of step 2.1 gives $1 / u_{j} = (ι (j + 1)^{- 1})^{1 / (j + 1)} \leq α_{j} \leq 1^{1 / (j + 1)} = 1$ .

step 2.1L2L3

4.1

Since $u_{j} \to 1$ with $u_{j} \geq 1 > 0$ , the quotient rule gives $1 / u_{j} \to 1$ ; so $α_{j} \to 1$ by the squeeze theorem, and therefore ${lim sup}_{j} α_{j} = 1$ .

step 3.1L1L4L5

5.1

Both families have root quantity exactly $1$ , yet one series diverges and the other converges; the claim is refuted, and the third clause of the root test is confirmed as unavoidable.

step 4.1step 2.2step 1.3step 1.4∎

Remarks

Every $p$ -series has root quantity $1$ . The computation in step 1.2 generalises verbatim: for rational $p > 0$ the root family of $k^{- p}$ is $u_{j}^{- p}$ , which tends to $1$ because $u_{j}$ does. So the root test is silent on the entire $p$ -series family, which is precisely the family the condensation test settles.
The root test and the ratio test are silent on the same family. The ratios of $k^{- p}$ also tend to $1$ , so neither test separates $p = 1 / 2$ from $p = 2$ . What does separate them is Raabe's test, whose expression reads the rate at which the ratios approach $1$ ; the companion example on this page carries the case $p = 2$ .
Why the two exponents are $- 1 / 2$ and $- 2$ rather than $- 1$ and $- 2$ . Taking the divergent witness with a fractional exponent makes the point that the failure is not about the harmonic series in particular: the root quantity is blind to the exponent altogether, and any pair straddling $p = 1$ would do.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Two series with $a_{k} \leq b_{k}$ for all $k$ , $\sum b_{k}$ convergent and $\sum a_{k}$ divergent, when the terms may be negative

Statement refuted

Refuted claim: if $a_{k} \leq b_{k}$ for every $k \in N$ and $\sum b_{k}$ converges, then $\sum a_{k}$ converges.

This is If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ with its nonnegativity hypothesis deleted, and deleting it destroys the theorem. Take

$a_{k} : = - \frac{1}{k + 1}, b_{k} : = 0 (k \in N) .$

Then $a_{k} < 0 = b_{k}$ for every $k$ ; the series $\sum b_{k}$ converges, with all partial sums equal to $0$ and sum $0$ ; and $\sum a_{k}$ diverges, being $- 1$ times the harmonic series (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping, Convergent series add and scale termwise).

What exactly fails. The proof of the comparison test bounds the partial sums of $\sum a_{k}$ above by those of $\sum b_{k}$ and then reads convergence off boundedness, and that last step is available only for a nonnegative series (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum). Here the partial sums of $\sum a_{k}$ are indeed bounded above, by $0$ ; they are unbounded below, and the theorem's conclusion fails for exactly that reason.

Facts & Assumptions

Given: The sequences $a_{k} : = - 1 / ι (k + 1)$ and $b_{k} : = 0$ for $k \in N$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals $ι (k + 1)$ are positive, so $1 / ι (k + 1) > 0$ and $a_{k} < 0$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

A finite sum of zeros is zero, being the scalar multiple of any finite sum by $0$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L3]

A constant sequence converges to its value (Limits and Cauchy sequences of reals).

[L4]

The harmonic series $\sum_{k \geq 1} 1 / k$ diverges, and it is the series of the sequence $j \mapsto 1 / ι (j + 1)$ (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

For $c \neq 0$ , $\sum c x_{k}$ converges if and only if $\sum x_{k}$ converges (Convergent series add and scale termwise).

[L6]

For a series of nonnegative terms, convergence is equivalent to boundedness above of the partial sums (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L7]

The refuted claim: $a_{k} \leq b_{k}$ for all $k$ and convergence of $\sum b_{k}$ imply convergence of $\sum a_{k}$ .

Counterexample

technique · direct

1.1

For every $k \in N$ , $a_{k} = - 1 / ι (k + 1) < 0 = b_{k}$ , so in particular $a_{k} \leq b_{k}$ .

givenL1

1.2

The partial sums of $\sum b_{k}$ are $\sum_{j < n} 0 = 0$ for every $n$ , a constant sequence, so $\sum b_{k}$ converges with sum $0$ .

givenL2L3

1.3

The sequence $(a_{k})$ is $(- 1)$ times the sequence $j \mapsto 1 / ι (j + 1)$ , whose series is the harmonic series and diverges; since $- 1 \neq 0$ , $\sum a_{k}$ diverges.

givenL4L5

2.1

So the hypotheses of the claim hold for this pair while its conclusion fails, and the claim is false.

step 1.1step 1.2step 1.3L7

3.1

The genuine comparison test is untouched: it requires $0 \leq a_{k}$ from some index on, and here $a_{k} < 0$ at every index.

step 1.1L6∎

Remarks

The witness is as degenerate as possible on purpose. Taking $b_{k} = 0$ removes every question about the dominating series and isolates the single point at issue: a series bounded above by a convergent one need not converge if it is free to run away downwards. Any negative divergent series would do; this one is the shortest to verify.
One-sided boundedness is not convergence. The partial sums here are $- \sum_{k = 1}^{n} 1 / k$ , bounded above by $0$ and unbounded below. For a nonnegative series that situation cannot arise, which is exactly the content of A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum and the reason the sign hypothesis appears in every comparison statement on the main page.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

A nonnegative non-monotone sequence for which $\sum a_{k}$ and $\sum 2^{k} a_{2^{k}}$ behave differently

Statement refuted

Refuted claim: for every family $(a_{k})_{k \geq 1}$ with $a_{k} \geq 0$ , $\sum_{k \geq 1} a_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} a_{2^{j}}$ converges.

This is For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges with its monotonicity hypothesis deleted. Let $P : = {2^{j} : j \in N}$ be the set of powers of $2$ and define, for naturals $k \geq 1$ ,

$a_{k} : = {\begin{cases} 0 & if k \in P, \\ 1 & if k \notin P . \end{cases}$

Every term is nonnegative, and the family is not monotone in either direction: $a_{1} = 0 < 1 = a_{3}$ and $a_{3} = 1 > 0 = a_{4}$ , since $1 = 2^{0}$ and $4 = 2^{2}$ belong to $P$ while $3$ does not.

The condensed series is $\sum_{j \geq 0} 2^{j} a_{2^{j}} = \sum_{j \geq 0} 0$ , which converges with sum $0$ . The original series $\sum_{k \geq 1} a_{k}$ diverges, because $a_{k} = 1$ at arbitrarily large indices, so its terms do not tend to $0$ (If a series converges then its terms tend to $0$ ).

Facts & Assumptions

Given: $P = {2^{j} : j \in N}$ and the family $a_{k}$ defined above for naturals $k \geq 1$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^{m}$ ).

[L1]

Powers of $2$ : $2^{j} \geq 1$ , $2^{j + 1} = 2 \cdot 2^{j}$ , and $j \mapsto 2^{j}$ is strictly increasing, since $2 > 1$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Canonical naturals are positive and strictly increasing).

[L2]

The naturals are discrete: no natural lies strictly between $n$ and $n + 1$ (Discreteness: $σ (n)$ is the immediate successor).

[L3]

The principle of induction (The principle of mathematical induction).

[L4]

A finite sum of zeros is zero, and a constant sequence converges to its value (Laws of finite sums and finite products, Finite sums and finite products, by recursion, Limits and Cauchy sequences of reals).

[L5]

A series whose terms do not converge to $0$ diverges (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L6]

Condensation requires the family to be nonnegative and nonincreasing (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L7]

The refuted claim: nonnegativity alone suffices for the condensation equivalence.

Counterexample

technique · direct

1.1

Every $a_{k}$ is $0$ or $1$ , hence nonnegative, so the family satisfies the hypothesis of the claim.

givenL7

1.2

The family is not monotone: $1 = 2^{0} \in P$ and $4 = 2^{2} \in P$ give $a_{1} = a_{4} = 0$ , while $3 \notin P$ gives $a_{3} = 1$ ; so $a_{1} < a_{3}$ rules out nonincreasing and $a_{3} > a_{4}$ rules out nondecreasing. That $3 \notin P$ holds because $2^{1} = 2 < 3 < 4 = 2^{2}$ and $j \mapsto 2^{j}$ is strictly increasing, so a power of $2$ equal to $3$ would force a natural strictly between $1$ and $2$ .

givenL1L2

1.3

Every condensed term is $2^{j} a_{2^{j}} = 2^{j} \cdot 0 = 0$ , since $2^{j} \in P$ for every $j$ .

givenL1

1.4

An induction gives $2^{n} > ι (n)$ for every $n \in N$ : at $n = 0$ this reads $1 > 0$ ; and if $2^{n} > ι (n)$ then $2^{n + 1} = 2^{n} + 2^{n} \geq 2^{n} + 1 > ι (n) + 1 = ι (n + 1)$ .

L1L3

1.5

For every $n \geq 1$ the natural $2^{n} + 1$ is not in $P$ : it satisfies $2^{n} < 2^{n} + 1 < 2^{n + 1}$ , the second inequality because $2^{n + 1} = 2^{n} + 2^{n} \geq 2^{n} + 2$ ; so a power of $2$ equal to it would force a natural strictly between $n$ and $n + 1$ .

L1L2

2.1

So the condensed series has all partial sums equal to $0$ and converges, with sum $0$ .

step 1.3L4

2.2

Hence for every $n \geq 1$ the index $k : = 2^{n} + 1$ satisfies $k > ι (n) \geq n$ and $a_{k} = 1$ , so $a_{k} = 1$ at indices exceeding any prescribed bound.

step 1.4step 1.5given

3.1

Therefore the terms of $\sum_{k \geq 1} a_{k}$ do not converge to $0$ : with the rational tolerance $1 / 2$ no index $K$ satisfies $∣ a_{k} ∣ < 1 / 2$ for all $k \geq K$ . So that series diverges.

step 2.2L5

4.1

The condensed series converges while the original diverges, so the claimed equivalence fails and the claim is false; the genuine condensation theorem is untouched, since its monotonicity hypothesis is violated here.

step 2.1step 3.1step 1.2L6L7∎

Remarks

The witness knocks out exactly one estimate. Condensation squeezes the block $a_{2^{n}}, \dots, a_{2^{n + 1} - 1}$ between $2^{n}$ copies of its last term and $2^{n}$ copies of its first, and both bounds are consequences of monotonicity. Here the first term of each block is $0$ and the rest are $1$ , so the upper bound $2^{n} a_{2^{n}} = 0$ is wildly wrong, and it is the upper bound that the convergence direction of the theorem uses.
The failure is one-directional here, and the other direction can fail too. This witness has a convergent condensed series and a divergent original. The complementary family $a_{k} : = 1 / k$ for $k \in P$ and $a_{k} : = 0$ otherwise reverses the roles, its original series being a geometric one and its condensed series having every term equal to $1$ ; that variant is not verified here, and only the direction exhibited above is claimed.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

With $a_{k} / b_{k} \to 0$ , convergence of $\sum a_{k}$ does not give convergence of $\sum b_{k}$

Statement refuted

Refuted claim: if $a_{k}, b_{k} > 0$ and $(a_{k} / b_{k})$ converges with $\lim_{k} a_{k} / b_{k} = 0$ , then convergence of $\sum a_{k}$ implies convergence of $\sum b_{k}$ .

Claim 2 of For $a_{k}, b_{k} > 0$ with $a_{k} / b_{k} \to L$ : if $L \in (0, \infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each gives the implication in the other direction only: at $L = 0$ , convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ . The claim above reverses it, and the reversal fails. Take

$a_{k} : = \frac{1}{(k + 1)^{2}}, b_{k} : = \frac{1}{k + 1} (k \in N) .$

Both are positive, and $a_{k} / b_{k} = 1 / (k + 1) \to 0$ . But $\sum a_{k}$ is $\sum_{k \geq 1} 1 / k^{2}$ , which converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ at $p = 2$ ), while $\sum b_{k}$ is the harmonic series, which diverges (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping).

The asymmetry is not an artefact of the proof. At $L = 0$ the hypothesis says the $a_{k}$ are eventually much smaller than the $b_{k}$ ; smallness of the $a_{k}$ can never constrain the $b_{k}$ from above, and the witness shows that it does not.

Facts & Assumptions

Given: The sequences $a_{k} : = 1 / ι (k + 1)^{2}$ and $b_{k} : = 1 / ι (k + 1)$ for $k \in N$ , and their quotients $q_{k} = a_{k} / b_{k}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^{m}$ , Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals $ι (k + 1)$ are positive, so $a_{k}, b_{k} > 0$ ; and reciprocation on the positives is order reversing (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

[L3]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; at $p = 2$ it converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Rational powers $a^{r}$ of a positive base).

[L4]

The harmonic series $\sum_{k \geq 1} 1 / k$ diverges, and it is the series of $j \mapsto 1 / ι (j + 1)$ (The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

Claim 2 of the limit comparison test: with $\lim_{k} a_{k} / b_{k} = 0$ , convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and that is the only implication it supplies in this regime (For $a_{k}, b_{k} > 0$ with $a_{k} / b_{k} \to L$ : if $L \in (0, \infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each).

[L6]

The refuted claim: with $\lim_{k} a_{k} / b_{k} = 0$ , convergence of $\sum a_{k}$ gives convergence of $\sum b_{k}$ .

Counterexample

technique · direct

1.1

Every $a_{k}$ and every $b_{k}$ is positive, so the quotients are defined and the hypotheses of the claim are available for this pair.

givenL1

1.2

The series $\sum a_{k}$ is $\sum_{k \geq 1} 1 / k^{2}$ , the $p$ -series at $p = 2$ , and it converges.

givenL3

1.3

The series $\sum b_{k}$ is the harmonic series, and it diverges.

givenL4

2.1

The quotients are $q_{k} = \frac{1 / ι (k + 1)^{2}}{1 / ι (k + 1)} = \frac{1}{ι (k + 1)}$ , and $(q_{k})$ converges to $0$ : given a rational $ε > 0$ , choose a natural $n \geq 1$ with $1 / n < ε$ , and then $0 < q_{k} \leq 1 / n < ε$ for every $k$ with $k + 1 \geq n$ .

step 1.1L1L2choose

3.1

So $\lim_{k} q_{k} = 0$ and $\sum a_{k}$ converges while $\sum b_{k}$ diverges; the claim is refuted.

step 2.1step 1.2step 1.3L6

4.1

Nothing in the limit comparison test is contradicted: its claim 2 asserts the implication in the opposite direction, and here its hypothesis, convergence of $\sum b_{k}$ , is false.

step 3.1L5∎

Remarks

The same pair also shows the divergence form is one-directional. Read contrapositively, claim 2 says divergence of $\sum a_{k}$ forces divergence of $\sum b_{k}$ . The witness has $\sum b_{k}$ divergent and $\sum a_{k}$ convergent, so divergence of the larger series says nothing about the smaller one, which is the same asymmetry seen from the other side.
The regime $L = + \infty$ fails symmetrically. Exchanging the roles of $(a_{k})$ and $(b_{k})$ in the witness gives $b_{k} / a_{k} \to + \infty$ with $\sum b_{k}$ divergent and $\sum a_{k}$ convergent, so claim 3 of the test is one-directional for the same reason. That reading is immediate from the computation above, the two sequences being the same two.

Series: Convergence and the Nonnegative Tests: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The harmonic series ∑1/k∑1/k diverges, by condensation and by Oresme block grouping

Example

Facts & Assumptions

Verification

Remarks

∑1/k2∑1/k2 converges with sum at most 22, by comparison with the telescoping ∑1/(k(k−1))∑1/(k(k−1))

Example

Facts & Assumptions

Verification

Remarks

Geometric sums computed: ∑k≥12−k=1∑k≥1​2−k=1 and ∑k≥0(−1/3)k=3/4∑k≥0​(−1/3)k=3/4

Example

Facts & Assumptions

Verification

Remarks

∑k≥11/(k(k+1))=1∑k≥1​1/(k(k+1))=1

Example

Facts & Assumptions

Verification

Remarks

Condensation reduces ∑1/kp∑1/kp to a geometric series with ratio 21−p21−p

Example

Facts & Assumptions

Verification

Remarks

Abel-Dini applied to ∑1/k∑1/k: ∑1/(ksk)∑1/(ksk​) still diverges while ∑1/(ksk2)∑1/(ksk2​) converges

Example

Facts & Assumptions

Verification

Remarks

A series with ratio limit exactly 11 that Raabe decides

Example

Facts & Assumptions

Verification

Remarks

ak=2−k+(−1)kak​=2−k+(−1)k has ratio limsup 22 and liminf 1/81/8, so the ratio test fails, while the root test gives convergence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

∑k−1/2∑k−1/2 diverges and ∑k−2∑k−2 converges, and both have root limit exactly 11

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Two series with ak≤bkak​≤bk​ for all kk, ∑bk∑bk​ convergent and ∑ak∑ak​ divergent, when the terms may be negative

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A nonnegative non-monotone sequence for which ∑ak∑ak​ and ∑2ka2k∑2ka2k​ behave differently

Statement refuted

Facts & Assumptions

Counterexample

Remarks

With ak/bk→0ak​/bk​→0, convergence of ∑ak∑ak​ does not give convergence of ∑bk∑bk​

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

The harmonic series $\sum 1 / k$ diverges, by condensation and by Oresme block grouping

$\sum 1 / k^{2}$ converges with sum at most $2$ , by comparison with the telescoping $\sum 1 / (k (k - 1))$

Geometric sums computed: $\sum_{k \geq 1} 2^{- k} = 1$ and $\sum_{k \geq 0} (- 1 / 3)^{k} = 3 / 4$

$\sum_{k \geq 1} 1 / (k (k + 1)) = 1$

Condensation reduces $\sum 1 / k^{p}$ to a geometric series with ratio $2^{1 - p}$

Abel-Dini applied to $\sum 1 / k$ : $\sum 1 / (k s_{k})$ still diverges while $\sum 1 / (k s_{k}^{2})$ converges

A series with ratio limit exactly $1$ that Raabe decides

$a_{k} = 2^{- k + (- 1)^{k}}$ has ratio limsup $2$ and liminf $1 / 8$ , so the ratio test fails, while the root test gives convergence

$\sum k^{- 1 / 2}$ diverges and $\sum k^{- 2}$ converges, and both have root limit exactly $1$

Two series with $a_{k} \leq b_{k}$ for all $k$ , $\sum b_{k}$ convergent and $\sum a_{k}$ divergent, when the terms may be negative

A nonnegative non-monotone sequence for which $\sum a_{k}$ and $\sum 2^{k} a_{2^{k}}$ behave differently

With $a_{k} / b_{k} \to 0$ , convergence of $\sum a_{k}$ does not give convergence of $\sum b_{k}$