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

Question Number 31979 by abdo imad last updated on 17/Mar/18

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

Commented by abdo imad last updated on 22/Mar/18

l e t p u t S_{n} = \sum_{k = 2}^{n} \frac{1}{{(k^{2} - 1)}^{2}}

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

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

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

d = {l i m}_{x \to - 1} {(x + 1)}^{2} F (x) = \frac{1}{4} w e h a v e F (- x) = F (x) \Rightarrow

\frac{- a}{x + 1} - \frac{c}{x - 1} + \frac{b}{{(x + 1)}^{2}} + \frac{d}{{(x - 1)}^{2}} = F (x) \Rightarrow c = - a \Rightarrow F (0)

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

F (0) = 1 = - 2 a + \frac{1}{2} \Rightarrow 2 a = - \frac{1}{2} \Rightarrow a = - \frac{1}{4} a n d

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

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

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

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

= - H_{n - 1} + H_{n + 1} - \frac{3}{2} + ξ_{n - 1} (2) + ξ_{n + 1} (2) - \frac{5}{4}

= H_{n + 1} - H_{n - 1} - \frac{11}{4} + ξ_{n - 1} (2) + ξ_{n + 1} (2) b u t

H_{n + 1} - H_{n - 1}_{n \to \infty} \to 0 ξ_{n - 1} (2) \to \frac{π^{2}}{6}

ξ_{n + 1} (2) \to \frac{π^{2}}{6} \Rightarrow 4 S_{n} \to \frac{π^{2}}{3} - \frac{11}{4} \Rightarrow

{l i m}_{n \to \infty} S_{n} = \frac{π^{2}}{12} - \frac{11}{16} . l e t r e m e m b e r t h a t H_{n} = \sum_{k = 1}^{n} \frac{1}{k}

a n d ξ_{n} (x) = \sum_{k = 1}^{n} \frac{1}{k^{x}} .