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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-27
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The Basel sum is pi squared over six by a residue computation

Statement

∑n=1∞1n2=π26.

This computation uses πcot⁡(πz)/z2 directly. It does not follow by substituting f(z)=1/z2 into the cotangent summation theorem, because that theorem excludes integer poles of f.

Facts & Assumptions

Given: The meromorphic function F(z)=πcot⁡(πz)/z2.

[L2]

The residue theorem applies on expanding rectangles, and the same boundary estimate as in the cotangent summation proof makes the rectangle integral of F tend to 0 (The residue theorem for a null-homologous cycle).

Proof

technique · direct
1.1L1algebra

At every nonzero integer n, the function πcot⁡(πz) has residue 1, so F has residue 1/n2 there.

1.2L1algebra

Near 0 one has sin⁡(πz)=πz−π3z36+O(z5),cos⁡(πz)=1−π2z22+O(z4), so πcot⁡(πz)=1z−π23z+O(z3). Therefore F(z)=1z3−π23z+O(z), and the residue of F at 0 is −π2/3.

2.1step 1.1step 1.2L2∎

Integrate F around the rectangles used in the cotangent theorem. By [L2], the boundary integral tends to 0, so the sum of the enclosed residues tends to 0. Hence 2∑n=1∞1n2−π23=0, which rearranges to the Basel value.

Depends on

Used by

Dependency tree · two levels

18 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