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Question Number 167820 by MWSuSon last updated on 26/Mar/22

given x^{2} = \frac{π^{2}}{3} + 4 Σ {(- 1)}^{n} \frac{\cos (nx)}{n^{2}}, show that Σ \frac{1}{{(2 n - 1)}^{2}} = \frac{π^{2}}{8}

Answered by Mathspace last updated on 26/Mar/22

x^{2} = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x)

x = \frac{π}{2} \Rightarrow \frac{π^{2}}{4} = \frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (\frac{n π}{2})

n = 2 m \Rightarrow c o s (n \frac{π}{2}) = {(- 1)}^{m}

n = 2 m + 1 \Rightarrow c o s (\frac{n π}{2}) = 0 \Rightarrow

4 \sum_{m = 1}^{\infty} \frac{{(- 1)}^{2 m} {(- 1)}^{m}}{4 m^{2}} = \frac{π^{2}}{4} - \frac{π^{2}}{3}

= - \frac{π^{2}}{12} \Rightarrow \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m}}{m^{2}} = - \frac{π^{2}}{12} \Rightarrow

\frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = - \frac{π^{2}}{12}

\Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{1}{4} \frac{π^{2}}{6} + \frac{π^{2}}{12}

= \frac{π^{2} + 2 π^{2}}{24} = \frac{3 π^{2}}{24} = \frac{π^{2}}{8}

n = r - 1 \Rightarrow \sum_{r = 1}^{\infty} \frac{1}{{(2 r - 1)}^{2}} = \frac{π^{2}}{8}