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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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]

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).

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

33 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)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

∑k≥11/(k(k+1))=1∑k≥1​1/(k(k+1))=1