CALCULUS-prove-that-n-1-1-n-1-n-1-2-2n-2log-2-golden-ratio-

Question Number 142970 by mnjuly1970 last updated on 08/Jun/21

\dots \dots \dots . C A L C U L U S \dots \dots \dots . .

p r o v e t h a t ::

ϕ := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} {((n - 1)!)}^{2}}{(2 n)!} = 2 {l o g}^{2} (φ)

φ = g o l d e n r a t i o \dots .

\dots \dots \dots \dots .

Answered by Dwaipayan Shikari last updated on 08/Jun/21

\underset{n = 1}{\sum^{\infty}} \frac{{((n - 1)!)}^{2}}{(2 n)!} {(- 1)}^{n - 1}

= \underset{n = 1}{\sum^{\infty}} \frac{Γ (n) Γ (n)}{2 n Γ (2 n)} {(- 1)}^{n - 1} = \frac{1}{2} \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n} x^{n - 1} {(1 - x)}^{n - 1} d x

= \frac{1}{2} \int_{0}^{1} \frac{1}{x (1 - x)} l o g (1 + x (1 - x)) d x

= \frac{1}{2} \int_{0}^{1} \frac{l o g (1 + x (1 - x))}{x} + \frac{l o g (1 + x (1 - x))}{1 - x} d x

= \int_{0}^{1} \frac{l o g (1 + x - x^{2})}{x} d x

= \int_{0}^{1} \frac{l o g (x)}{x - \frac{1}{φ}} + \frac{l o g (x)}{x + φ} d x = - \frac{1}{φ} \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} {(x φ)}^{n} l o g (x) + \frac{1}{φ} \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} {(- x φ)}^{n} l o g (x) d x

= \frac{1}{φ} {\underset{n = 1}{\sum^{\infty}} \frac{φ^{n}}{{(n + 1)}^{2}} + \frac{1}{φ} \underset{n = 1}{\sum^{\infty}} \frac{{(- φ)}^{n}}{{(n + 1)}^{2}}}

= \frac{1}{φ^{2}} ({L i}_{2} (φ) - φ) - \frac{1}{φ^{2}} ({L i}_{2} (- φ) + φ)

= \frac{1}{φ^{2}} ({L i}_{2} (φ) - {L i}_{2} (- φ))