ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$\sum_{k \ge 1} 1/(k(k+1)) = 1$

Example

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

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 $\sum_{k \ge 1} 1/k$ , 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 $b_j := 1/\iota(j+1)$ for $j \in \mathbb{N}$ , so that $b_j = 1/k$ at $k = j+1$ ; and the family $d_k := 1/(k(k+1))$ for naturals $k \ge 1$ , so that $\sum_{k \ge 1} d_k$ 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 $\iota(j+1)$ 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 $\varepsilon > 0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Limits and Cauchy sequences of reals).

Verification

technique · direct

1.1

For every $j \in \mathbb{N}$ : $b_j - b_{j+1} = \dfrac{1}{\iota(j+1)} - \dfrac{1}{\iota(j+2)} = \dfrac{\iota(j+2) - \iota(j+1)}{\iota(j+1)\iota(j+2)} = \dfrac{1}{\iota(j+1)\,\iota(j+2)}$ .

givenL1algebra

1.2

The sequence $(b_j)$ converges to $0$ : given a rational $\varepsilon > 0$ , choose $n \ge 1$ with $1/n < \varepsilon$ ; then for every $j$ with $j+1 \ge n$ we have $0 < b_j \le 1/n < \varepsilon$ .

givenL1L3choose

2.1

The term of $\sum_{k \ge 1} d_k$ at index $j$ is $d_{j+1} = \dfrac{1}{\iota(j+1)\,\iota(j+2)}$ , 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 $b_0 - 0 = 1/\iota(1) = 1$ .

step 1.2L2L1

3.1

Therefore $\sum_{k \ge 1} 1/(k(k+1))$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 68 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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\sum_{k \ge 1} 1/(k(k+1)) = 1∑k≥1​1/(k(k+1))=1