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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,RN12N+3,

where

RN:=(1)N+101x2N+21+x2dx.

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, tant=sint/cost, sect=1/cost, and (tant)=sec2t; for every real t, sin2t+cos2t=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 sin0=0, cos0=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 M1 with 1/M<ε (For every ε>0 in a complete ordered field there is a natural n1 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 cos2(π/4)=sin2(π/4), while the stated monotonicities make both values positive; hence [L6] gives tan(π/4)=1, and [L6] also gives tan0=0. The definitions and Pythagorean identity in [L6] give sec2t=1+tan2t. Apply [L5] with x=tant on [0,π/4]. Then dx=(1+tan2t)dt, so 01dx1+x2=0π/41dt=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+101x2N+2/(1+x2)dx.

step 1.1L2L3
3.1

On [0,1], 0x2N+2/(1+x2)x2N+2, so [L3] and [L4] give RN1/(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

Depends on

Used by

Dependency tree · next 3 levels

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