find-the-value-of-n-0-1-n-2n-1-2n-3-

Question Number 31982 by abdo imad last updated on 17/Mar/18

f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 3)} .

Commented by 6123 last updated on 18/Mar/18

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 3)} = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\frac{1}{2 n + 1} - \frac{1}{2 n + 3}) .

= \frac{1}{2} [(1 - \frac{1}{3}) + (\frac{1}{5} - \frac{1}{3}) + (\frac{1}{5} - \frac{1}{7}) + (\frac{1}{9} - \frac{1}{7}) + \dots]

= \frac{1}{2} [1 + 2 (\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \dots) - 2 (\frac{1}{3} + \frac{1}{7} + \frac{1}{11} + \dots)]

t h e r e s t i s h i s t o r y .

Commented by abdo imad last updated on 18/Mar/18

l e t p u t S = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) (2 n + 3)} a n d

S (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{(2 n + 1) (2 n + 3)} w e h a v e S (- 1) = S b u t

S (x) = \frac{1}{2} (\sum_{n = 0}^{\infty} (\frac{1}{2 n + 1} - \frac{1}{2 n + 3}) x^{n}) \Rightarrow f o r x \neq o

2 S (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} - \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 3} b u t

\sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 3} = \frac{1}{3} + \sum_{n = 1}^{\infty} \frac{x^{n}}{2 n + 3} = \frac{1}{3} + \sum_{n = 0}^{\infty} \frac{x^{n - 1}}{2 n + 1}

= \frac{1}{3} + \frac{1}{x} \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} \Rightarrow

2 S (x) = - \frac{1}{3} + (1 - \frac{1}{x}) \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} l e t c a l c u l a t e f o r - 1 ⩽ x < 0

A (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} \Rightarrow A (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(- x)}^{n}}{2 n + 1}

= \frac{1}{\sqrt{- x}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(\sqrt{- x})}^{2 n + 1}}{2 n + 1} = \frac{1}{\sqrt{- x}} φ (\sqrt{- x}) w i t h

φ (t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} t^{2 n + 1}}{2 n + 1} \Rightarrow φ^{^{'}} (t) = \sum_{n = 0}^{\infty} {(- t^{2})}^{n} = \frac{1}{1 + t^{2}}

\Rightarrow φ (t) = a r c t a n t + λ b u t λ = φ (0) = 0 \Rightarrow

A (x) = \frac{1}{\sqrt{- x}} r c t a n (\sqrt{- x}) s o f o r - 1 ⩽ x < 0 w e h a v e

S (x) = - \frac{1}{6} + \frac{1}{2} \frac{x - 1}{x} \frac{1}{\sqrt{- x}} a r c t a n (\sqrt{- x})

= \frac{x - 1}{2 x \sqrt{- x}} a r c t a n (\sqrt{- x}) - \frac{1}{6} a n d

S (- 1) = S = \frac{- 2}{- 2} a r c t a n (1) - \frac{1}{6} = \frac{π}{4} - \frac{1}{6} .