Find-n-1-1-2n-1-4-

Question Number 1539 by 314159 last updated on 17/Aug/15

Find \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n - 1)}^{4}} .

Answered by 123456 last updated on 17/Aug/15

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{4}} = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(2 n - 1)}^{4}} + \underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(2 n)}^{4}}

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

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

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(2 n - 1)}^{4}} = \frac{15}{16} \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{4}}

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(2 n - 1)}^{4}} = \frac{15}{16} \times \frac{π^{4}}{90} = \frac{15 π^{4}}{16 \times 90} = \frac{5 π^{4}}{16 \times 30} = \frac{π^{4}}{16 \times 6} = \frac{π^{4}}{96}

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(2 n - 1)}^{4}} = \frac{π^{4}}{96}

Commented by 314159 last updated on 17/Aug/15

Thank a lot \dots

Commented by 112358 last updated on 17/Aug/15

O u t o f c u r i o s i t y, h o w i s

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{4}} = \frac{π^{4}}{90} ?

Commented by 123456 last updated on 17/Aug/15

rieman zeta function, later i answer you

Commented by 112358 last updated on 17/Aug/15

T h a n k s

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