find-the-value-of-the-sum-n-1-1-2n-1-2-2n-1-2-

Question Number 36819 by maxmathsup by imad last updated on 06/Jun/18

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

Commented by maxmathsup by imad last updated on 03/Aug/18

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

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

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

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

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

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

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

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

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

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

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

= u_{n} - u_{0} = \frac{1}{2 n + 1} - 1 a l s o

\sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} =_{k = j + 1} \sum_{j = 0}^{n - 1} \frac{1}{{(2 j + 1)}^{2}} \to \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

\sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}} = \sum_{k = 0}^{n} \frac{1}{{(2 k + 1)}^{2}} - 1 \to \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} - 1

w e h a v e \frac{π^{2}}{6} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{1}{4} \frac{π^{2}}{6} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

\Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{6} - \frac{π^{2}}{24} = \frac{3 π^{2}}{24} = \frac{π^{2}}{8} \Rightarrow

{l i m}_{} S_{n} = \frac{1}{4} {- 1 + \frac{π^{2}}{8} + \frac{π^{2}}{8} - 1} = \frac{1}{4} {\frac{π^{2}}{4} - 2} = \frac{π^{2}}{16} - \frac{1}{2}

S = \frac{π^{2}}{16} - \frac{1}{2} .