calculate-n-1-1-n-2n-1-n-2-

Question Number 72889 by mathmax by abdo last updated on 04/Nov/19

c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1) n^{2}}

Commented by mathmax by abdo last updated on 04/Nov/19

l e t S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} (2 n + 1)} f i r s t l e t d e c o m p o s e F (x) = \frac{1}{x^{2} (2 x + 1)}

F (x) = \frac{a}{x} + \frac{b}{x^{2}} + \frac{c}{2 x + 1}

b = {l i m}_{x \to 0} x^{2} F (x) = 1

c = {l i m}_{x \to - \frac{1}{2}} (2 x + 1) F (x) = 4 \Rightarrow F (x) = \frac{a}{x} + \frac{1}{x^{2}} + \frac{4}{2 x + 1}

{l i m}_{x \to + \infty} x F (x) = 0 = a + \frac{c}{2} \Rightarrow a = - 2 \Rightarrow F (x) = - \frac{2}{x} + \frac{1}{x^{2}} + \frac{4}{2 x + 1} \Rightarrow

S = \sum_{n = 1}^{\infty} {(- 1)}^{n} F (n) = - 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} = - l n (2)

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = (2^{1 - 2} - 1) ξ (2) = - \frac{π^{2}}{12}

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} - 1 = \frac{π}{4} - 1 \Rightarrow

S = - 2 (- l n 2) - \frac{π^{2}}{12} + 4 (\frac{π}{4} - 1) = 2 l n (2) - \frac{π^{2}}{12} + π - 4

Answered by mind is power last updated on 04/Nov/19

\frac{1}{(2 n + 1) n^{2}} = \frac{1}{n^{2}} + \frac{c}{n} + \frac{4}{2 n + 1} \Rightarrow \frac{1}{n^{2}} + \frac{- 2}{n^{}} + \frac{4}{2 n + 1}

\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n^{2}} + \sum_{n ⩾ 1} \frac{{(- 1)}^{n + 1}}{n} + 4 Σ \frac{{(- 1)}^{n}}{2 n + 1}

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

\arctan (x) = Σ {(- 1)}^{n} . \frac{x^{2 n + 1}}{2 n + 1}

x = 1 \Rightarrow Σ \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4} - 1

\sum_{n ⩾ 1} \frac{{(- 1)}^{n + 1}}{n} \to \ln (2)

\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{4} . ζ (2) - \frac{3}{4} ζ (2) = - \frac{ζ (2)}{2} = - \frac{π^{2}}{12}

\Rightarrow \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{(2 n + 1) n^{2}} = - \frac{π^{2}}{12} + 2 \ln (2) + π - 4

Commented by turbo msup by abdo last updated on 04/Nov/19

t h s n k y o u s i r .

Commented by mind is power last updated on 04/Nov/19

y, re welcom