ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

56 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Telescoping series (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

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