CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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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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The partial-fraction expansion of pi-squared cosecant-squared

Statement

For every $z \in C ∖ Z$ ,

$π^{2} \csc^{2} (π z) = \sum_{n \in Z} \frac{1}{(z - n)^{2}} = \frac{1}{z^{2}} + \sum_{n \geq 1} (\frac{1}{(z - n)^{2}} + \frac{1}{(z + n)^{2}}) .$

with locally uniform convergence on $C ∖ Z$ .

Facts & Assumptions

Given: The cotangent expansion on $C ∖ Z$ .

[L1]

On $C ∖ Z$ , $π \cot (π z) = 1 / z + \sum_{n \geq 1} 2 z / (z^{2} - n^{2})$ (The Mittag-Leffler expansion of pi cotangent).

[L2]

The derivative of $\cot$ is $- \csc^{2}$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives).

[L3]

Locally uniform convergence of holomorphic partial sums forces local-uniform convergence of their derivatives to the derivative of the limit (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

Proof

technique · direct

1.1L1L3given

Let $S_{N} (z) : = \frac{1}{z} + \sum_{n = 1}^{N} (\frac{1}{z - n} + \frac{1}{z + n}) .$ By [L1], the holomorphic functions $S_{N}$ converge locally uniformly on $C ∖ Z$ to $π \cot (π z)$ . On any compact set $K \subseteq C ∖ Z$ , the derivatives satisfy $S_{N}^{'} (z) = - \frac{1}{z^{2}} - \sum_{n = 1}^{N} (\frac{1}{(z - n)^{2}} + \frac{1}{(z + n)^{2}}),$ and this derivative series converges uniformly on $K$ because its terms are $O (n^{- 2})$ there. Therefore [L3] identifies the derivative of the limit with the displayed series on $K$ .

2.1step 1.1L2algebra∎

Differentiating [L1] gives $- π^{2} \csc^{2} (π z) = - \frac{1}{z^{2}} - \sum_{n \geq 1} (\frac{1}{(z - n)^{2}} + \frac{1}{(z + n)^{2}})$ by [L2]. Multiplying by $- 1$ yields the claimed formula.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

M. Weber, Complex Analysis, Example 3.3.1 (standard reference, not scraped)
J. Lebl, Guide to Cultivating Complex Analysis, §9.4 (standard reference, not scraped)