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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...

Statement

The series

∑k=0∞(−1)k2k+1

converges, and its sum is π/4. More precisely, for every natural N,

π4=∑k=0N(−1)k2k+1+RN,∣RN∣≤12N+3,

where

RN:=(−1)N+1∫01x2N+21+x2 dx.

Facts & Assumptions

Given: A natural N and the finite geometric identity used below.

[L1]

A series converges exactly when its sequence of finite partial sums converges (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

On its natural domain, tan⁡t=sin⁡t/cos⁡t, sec⁡t=1/cos⁡t, and (tan⁡t)′=sec⁡2t; for every real t, sin⁡2t+cos⁡2t=1 (Tangent, cotangent, secant, and cosecant on their exact natural domains, Derivatives and fundamental periods of tangent, cotangent, secant, and cosecant, Parity and the Pythagorean identity for sine and cosine).

[L7]

The quarter-turn values are sin⁡(π/2)=1 and cos⁡(π/2)=0; the sine and cosine addition formulas hold for all real inputs, sine is strictly increasing on [−π/2,π/2], cosine is strictly decreasing on [0,π], and sin⁡0=0, cos⁡0=1 (Quarter-turn values and shifts by pi/2 and pi, The addition formulas for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine, The derivatives of sine and cosine are cosine and minus sine).

[L8]

A sequence squeezed between two sequences with the same limit has that limit (The squeeze theorem).

[L9]

For every ε>0 there is a natural M≥1 with 1/M<ε (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

Proof

technique · direct
1.1

For every real x, finite geometric algebra gives 11+x2=∑k=0N(−1)kx2k+(−1)N+1x2N+21+x2.

givenalgebra
1.2

From [L7], the cosine double-angle formula at π/4 gives cos⁡2(π/4)=sin⁡2(π/4), while the stated monotonicities make both values positive; hence [L6] gives tan⁡(π/4)=1, and [L6] also gives tan⁡0=0. The definitions and Pythagorean identity in [L6] give sec⁡2t=1+tan⁡2t. Apply [L5] with x=tan⁡t on [0,π/4]. Then dx=(1+tan⁡2t)dt, so ∫01dx1+x2=∫0π/41 dt=π4.

L4L5L6L7algebra
2.1

Integrating step 1.1 on [0,1] and using [L2] and [L3] yields ∫01dx1+x2=∑k=0N(−1)k2k+1+RN, where RN=(−1)N+1∫01x2N+2/(1+x2) dx.

step 1.1L2L3
3.1

On [0,1], 0≤x2N+2/(1+x2)≤x2N+2, so [L3] and [L4] give ∣RN∣≤1/(2N+3).

step 2.1L3L4algebra
4.1

Steps 2.1 and 1.2 give the displayed finite-remainder identity. By [L9], 1/(2N+3)→0, so step 3.1 and [L8] make the finite sums converge to π/4; by [L1], this is the sum of the series.

step 2.1step 3.1step 1.2L1L8L9∎

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