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

Question Number 38518 by math khazana by abdo last updated on 26/Jun/18

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

Commented by prof Abdo imad last updated on 27/Jun/18

l e t p u t S_{n} = \sum_{k = 1}^{n} \frac{1}{k^{2} {(2 k - 1)}^{2}} a n d l e t d e c o m p o s e

F (x) = \frac{1}{x^{2} {(2 x - 1)}^{2}}

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

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

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

F (x) = \frac{a}{x} + \frac{1}{x^{2}} + \frac{c}{2 x - 1} + \frac{4}{{(2 x - 1)}^{2}}

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

F (x) = \frac{a}{x} - \frac{2 a}{2 x - 1} + \frac{1}{x^{2}} + \frac{4}{{(2 x - 1)}^{2}}

F (1) = 1 = a - 2 a + 1 + 4 \Rightarrow - a + 4 = 0 \Rightarrow a = 4

F (x) = \frac{4}{x} - \frac{8}{2 x - 1} + \frac{1}{x^{2}} + \frac{4}{{(2 x - 1)}^{2}}

S_{n} = \sum_{k = 1}^{n} F (k) = 4 \sum_{k = 1}^{n} \frac{1}{k} - 8 \sum_{k = 1}^{n} \frac{1}{2 k - 1}

+ \sum_{k = 1}^{n} \frac{1}{k^{2}} + 4 \sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} b u t

\sum_{k = 1}^{n} \frac{1}{k} = H_{n}

\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \frac{1}{5} + \dots . + \frac{1}{2 n - 1}

= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots . . + \frac{1}{2 n - 1} + \frac{1}{2 n}

- \frac{1}{2} H_{n} = H_{2 n} - \frac{1}{2} H_{n}

\sum_{k = 1}^{n} \frac{1}{k^{2}} = ξ_{n} (2)

\sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} = \sum_{k = 2}^{n + 1} \frac{1}{{(2 k + 1)}^{2}}

= \sum_{k = 0}^{n + 1} \frac{1}{{(2 k + 1)}^{2}} \Rightarrow

S_{n} = 4 H_{n} - 8 H_{2 n} + 4 H_{n} + ξ_{n} (2) + 4 \sum_{k = 0}^{n + 1} \frac{1}{{(2 k + 1)}^{2}}

S_{n} = - 8 (H_{2 n} - H_{n}) + ξ_{n} (2) + 4 \sum_{k = 0}^{n + 1} \frac{1}{{(2 k + 1)}^{2}}

{l i m}_{n \to + \infty} S_{n} = - 8 l n (2) + \frac{π^{2}}{6} + 4 . \frac{π^{2}}{8}

= - 8 l n (2) + \frac{π^{2}}{6} + \frac{π^{2}}{2} = \frac{4 π^{2}}{6} - 8 l n (2)

= \frac{2 π^{2}}{3} - 8 l n (2) .