find-n-0-1-2n-1-4-

Question Number 45080 by turbo msup by abdo last updated on 08/Oct/18

f i n d \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{4}}

Commented by maxmathsup by imad last updated on 08/Oct/18

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

= \frac{1}{16} \sum_{n = 1}^{\infty} \frac{1}{n^{4}} + \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{4}} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{4}} = (1 - \frac{1}{16}) \sum_{n = 1}^{\infty} \frac{1}{n^{4}}

= \frac{15}{16} . \frac{π^{4}}{90} = \frac{15 π^{4}}{16 . 2 . 15.3} = \frac{π^{4}}{32.3} = \frac{π^{4}}{96}

★ \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{4}} = \frac{π^{4}}{96} ★