Series: Convergence and the Nonnegative Tests

Throughout, $\mathbb{R}$ is the complete ordered field (The real numbers) and a sequence of reals is a function $a : \mathbb{N} \to \mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), written $(a_k)$; recall that $\mathbb{N}$ contains $0$.

Partial sums. Let $(a_k)$ be a sequence of reals. Its sequence of partial sums is

$s_n \;:=\; \sum_{k < n} a_k \qquad (n \in \mathbb{N}),$

the finite sum of Finite sums and finite products, by recursion. In particular $s_0 = 0$, the empty sum, and $s_{n+1} = s_n + a_n$ for every $n$, those being exactly the two recursion clauses that define the finite sum. Note that $s_n$ is the sum of the $n$ terms $a_0, \dots, a_{n-1}$, so the index $n$ counts terms rather than naming the last one.

Convergence, the sum, divergence. The series of $(a_k)$, written $\sum a_k$, converges when the sequence $(s_n)$ of partial sums converges (Limits and Cauchy sequences of reals), and then the sum of the series is

$\sum_{k=0}^{\infty} a_k \;:=\; \lim_{n} s_n .$

The series diverges when $(s_n)$ does not converge. A convergent sequence of reals has exactly one limit (A sequence has at most one limit), so the displayed symbol names a single real number and nothing further has to be checked for it to be well defined.

Series with a general starting index. Let $m \in \mathbb{N}$ and let $a$ be a function on $\{\, k \in \mathbb{N} : k \ge m \,\}$, which we call a family from $m$ and write $(a_k)_{k \ge m}$. The series

$\sum_{k \ge m} a_k$

is by definition the series of the sequence $b_j := a_{j + m}$, $j \in \mathbb{N}$, which is a genuine sequence of reals; it converges exactly when that series converges, and its sum is then written $\sum_{k = m}^{\infty} a_k$. Its partial sums are

$\sum_{k=m}^{n-1} a_k \;=\; \sum_{j < n-m} a_{m+j} \qquad (n \ge m),$

in the notation of Finite sums and finite products, by recursion, the value at $n = m$ being the empty sum $0$. A sequence on $\mathbb{N}$ is the case $m = 0$, and the two readings of $\sum a_k$ agree there, since $b_j = a_j$.

This clause is not a convenience. Sequences in this library are functions on $\mathbb{N}$ and $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), while many of the classical series are built from expressions that are undefined at the index $0$: $1/k$, $1/k^{p}$ and $|a_k|^{1/k}$ all require $k \ge 1$. Writing such a series as $\sum_{k \ge 1} a_k$ names an honest object, whereas writing it as a sequence on $\mathbb{N}$ would require a value at an index where the defining expression has none. Every statement on this page says which starting index it uses.

Tail series. For $N \in \mathbb{N}$, the $N$-th tail series of $\sum a_k$ is $\sum_{k \ge N} a_k$, that is the series of the $N$-th tail $a^{(N)}$ of Sequences of reals: bounded, eventually, frequently, tails, subsequences, whose terms are $a^{(N)}_j = a_{j+N}$. The $0$-th tail series is the series itself.

• "Diverges" here means "does not converge", and nothing more. A divergent series may have partial sums that run away to $+\infty$, or to $-\infty$, or that oscillate without settling anywhere. The three behaviours are not distinguished by the word, and no statement on this page uses "diverges" to mean "the partial sums are unbounded" unless it says so.

• The symbol $\sum_{k=0}^{\infty} a_k$ is defined only for a convergent series. It denotes a real number, not a formal object, and it is illegitimate to write it down before convergence has been established. Where a proof needs to speak of the series without knowing whether it converges, it speaks of $(a_k)$ and of $(s_n)$.

• Two indices, doing different work. The index $k$ runs over the terms and is bound; the index $n$ runs over the partial sums and is the variable in which the limit is taken. Confusing them is the commonest slip in the subject, and it is the reason the definition above fixes $s_n = \sum_{k<n} a_k$ rather than $\sum_{k \le n} a_k$: with this choice the recursion $s_{n+1} = s_n + a_n$ is the one supplied by Finite sums and finite products, by recursion, with no shift anywhere.

Let $(a_k)$ be a sequence of reals with partial sums $s_n = \sum_{k<n} a_k$, let $N \in \mathbb{N}$, and let $t_j := \sum_{i<j} a_{N+i}$ be the partial sums of the $N$-th tail series $\sum_{k \ge N} a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

1. $t_j = s_{j+N} - s_N$ for every $j \in \mathbb{N}$;
2. $\sum a_k$ converges if and only if its $N$-th tail series converges, and in that case $\sum_{k=0}^{\infty} a_k \;=\; s_N \;+\; \sum_{k=N}^{\infty} a_k ;$
3. hence the following are equivalent: $\sum a_k$ converges; every tail series of $\sum a_k$ converges; some tail series of $\sum a_k$ converges.

In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.

• Where the hypothesis-free character comes from. Nothing here assumes the terms have a sign or that any series converges: claim 1 is an identity between finite sums, valid always, and claims 2 and 3 are read off from it by two standard limit rules. That is why the lemma may be used to move between a series and its tails inside proofs that have not yet decided the convergence question.

• The sum does change. Only the fact of convergence is tail invariant. The identity in claim 2 is the exact bookkeeping: discarding the first $N$ terms lowers the sum by $s_N$, and no more.

Let $(a_k)$ be a sequence of reals. If the series $\sum a_k$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the sequence $(a_k)$ converges (Limits and Cauchy sequences of reals), with

$\lim_{k} a_k \;=\; 0 .$

Equivalently, in the contrapositive form in which the lemma is almost always used: if $(a_k)$ does not converge to $0$ then $\sum a_k$ diverges.

The same statement holds for a series $\sum_{k \ge m} a_k$ with a general starting index, with the conclusion that $(a_{j+m})_{j \in \mathbb{N}}$ converges to $0$: that series is by definition the series of the sequence $b_j = a_{j+m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series), so it is the statement above applied to $(b_j)$.

The converse is false; it is refuted by FALSE: if $a_k \to 0$ then $\sum a_k$ converges.

• This is a necessary condition and never a sufficient one. It rules a series out; it never rules one in. The harmonic series has terms tending to $0$ and diverges, which is exactly the content of FALSE: if $a_k \to 0$ then $\sum a_k$ converges.

• What the proof actually uses. Only that the partial sums converge and that consecutive partial sums differ by a term. No sign hypothesis is placed on $(a_k)$, and none is available at this point on the page.

Let $(a_k)$ be a sequence of reals, with partial sums $s_n = \sum_{k<n} a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then $\sum a_k$ converges if and only if

$\text{for every real } \varepsilon > 0 \text{ there is } N \in \mathbb{N} \text{ such that } \Big| \sum_{k=m+1}^{n} a_k \Big| < \varepsilon \text{ for all } n > m \ge N .$

The block $\sum_{k=m+1}^{n} a_k$ is the finite sum $a_{m+1} + \dots + a_n$ of Finite sums and finite products, by recursion, and it equals $s_{n+1} - s_{m+1}$.

This is the Cauchy criterion transported from sequences to series. Its value is that it decides convergence without producing, or even naming, the sum.

• The criterion is stated over blocks, not over partial sums, on purpose. In applications one estimates a run of consecutive terms $a_{m+1}, \dots, a_n$ directly; the translation into $|s_{n+1} - s_{m+1}|$ is step 1.1 and is done once here so that no later proof has to repeat it.

• Taking $n = m+1$ recovers the term test. The single-term block gives $|a_{m+1}| < \varepsilon$ for all $m \ge N$, which is $a_k \to 0$; so If a series converges then its terms tend to $0$ is the weakest consequence of this criterion. The criterion is strictly stronger, since it constrains arbitrarily long blocks and not only single terms.

Let $(a_k)$ and $(b_k)$ be sequences of reals whose series converge (Series, partial sums, convergence and the sum, divergence, and the tail series), and let $c \in \mathbb{R}$. Then:

1. $\sum (a_k + b_k)$ converges, with $\displaystyle\sum_{k=0}^{\infty}(a_k + b_k) = \sum_{k=0}^{\infty} a_k + \sum_{k=0}^{\infty} b_k$;
2. $\sum c\,a_k$ converges, with $\displaystyle\sum_{k=0}^{\infty} c\,a_k = c \sum_{k=0}^{\infty} a_k$.

Moreover, for $c \ne 0$ and an arbitrary sequence $(a_k)$, whose series is not assumed to converge:

3. $\sum c\,a_k$ converges if and only if $\sum a_k$ converges. Equivalently, $\sum a_k$ diverges if and only if $\sum c\,a_k$ diverges.

Claim 3 is the form used whenever a comparison is made against a constant multiple of a known series.

• There is no product rule here, and there is no rule for $\sum a_k b_k$. The proof works because a finite sum is additive and homogeneous, and neither property has an analogue for products. Multiplying series is a genuinely harder question, requiring absolute convergence, and it is not treated on this page.

• Claim 3 needs $c \ne 0$ and nothing else. In particular it does not need either series to converge, which is what makes it usable in the divergence direction: scaling a divergent series by a nonzero constant leaves it divergent.

Let $(a_k)$ be a sequence of reals. If the series $\sum |a_k|$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the series $\sum a_k$ converges.

A series with the property that $\sum |a_k|$ converges is called absolutely convergent; the lemma says that absolute convergence implies convergence.

The same statement holds for a family from a general starting index $m$, being this statement applied to the shifted sequence $j \mapsto a_{j+m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

The converse is false, and the standard witness is the alternating harmonic series. That witness is not available on this page: its convergence is the alternating series test, which is not proved here. Nothing below asserts a converse, and no item on this page uses one.

• Nothing here identifies the two sums, and they are in general different. What is proved is that the second series converges, not that it converges to the same value; the bound $\big|\sum_{k=0}^{\infty} a_k\big| \le \sum_{k=0}^{\infty}|a_k|$ is true and is not needed anywhere on this page, so it is not proved here.

• Why the Cauchy criterion is the right instrument. The terms have no sign, so A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum does not apply to $\sum a_k$ and boundedness of its partial sums would prove nothing. The Cauchy criterion is the one convergence test on this page that never names a candidate sum and never asks for a sign, and the whole proof is the observation that its hypothesis for $(|a_k|)$ implies its hypothesis for $(a_k)$, term by term, through one application of the finite triangle inequality.

• What this unlocks on this page. The root test (Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing) and the ratio test (Ratio test: $\limsup |a_{k+1}/a_k| < 1$ gives absolute convergence and hence convergence, and $\liminf |a_{k+1}/a_k| > 1$ gives divergence) each produce convergence of $\sum |a_k|$ directly, by comparison with a geometric series; with this lemma both reach their standard conclusion, the convergence of $\sum a_k$ itself. Without it their convergence halves would be strictly weaker than the classical statements.

• The systematic theory is elsewhere. Rearrangement, the Riemann series theorem, conditional convergence and products of series all belong with absolute convergence and are developed on a later page of this track. This lemma is only the one implication those two tests need.

Let $r \in \mathbb{R}$ and let $r^k$ be the integer power (Integer powers $a^m$), so that $r^0 = 1$ for every $r$, including $r = 0$.

1. If $|r| < 1$ then the series $\sum r^k$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $\sum_{k=0}^{\infty} r^{k} \;=\; \frac{1}{1-r} .$
2. If $|r| \ge 1$ then $\sum r^k$ diverges.

The series starts at $k = 0$ and its first term is $r^0 = 1$; in particular $\sum_{k=0}^{\infty} 2^{-k} = 2$, while the series starting at $k = 1$ sums to $1$. Which starting index is meant has to be said, and it is said here.