Question-128767

Question Number 128767 by LUFFY last updated on 10/Jan/21

Commented by LUFFY last updated on 10/Jan/21

ans \frac{1}{24} but i am getting \frac{1}{12}

Commented by LUFFY last updated on 10/Jan/21

s e n d m e p l e a s e

Answered by Dwaipayan Shikari last updated on 10/Jan/21

\underset{n = 1}{\sum^{\infty}} \frac{n^{k}}{e^{2 π n} - 1} = \frac{k!}{{(2 π)}^{k + 1}} . ζ (k + 1) (k = 4 m + 1)

k = 13 I t i s \frac{13!}{{(2 π)}^{14}} . \frac{{(2 π)}^{14} B_{14}}{2 (14)!} = \frac{B_{14}}{28} = \frac{1}{24}

B_{n} = B e r n o u l l i N u m b e r

Commented by LUFFY last updated on 10/Jan/21

check again

Commented by LUFFY last updated on 10/Jan/21

can you send full solution

Commented by Dwaipayan Shikari last updated on 19/Feb/21

\frac{1}{e^{2 π} - 1} = \underset{n = 1}{\sum^{\infty}} e^{- 2 π n}

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + x^{2}} = \frac{π}{2 x} c o t h (π x) - \frac{1}{2 x^{2}}

\frac{1}{x} + 2 x \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + x^{2}} = π c o t h (π x) \Rightarrow \frac{1}{2 x} + x \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + x^{2}} = \frac{π}{2} + \frac{π}{e^{2 π x} - 1}

\frac{1}{e^{2 π x} - 1} = \frac{x}{π} \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + x^{2}} + \frac{1}{2 π x} - \frac{1}{2}

\underset{n = 1}{\sum^{\infty}} e^{- 2 π n x} = \frac{1}{2 π x} \underset{n = 1}{\sum^{\infty}} \frac{1}{x + i n} + \frac{1}{x - i n} + \frac{1}{2 π x} - \frac{1}{2}

d i f f e r e n t i a t i n g k t h t i m e (k = 4 m + 1)

{(2 π)}^{k} \underset{x = 1}{\sum^{\infty}} n^{k} e^{- 2 π n x} = \frac{k!}{{(x)}^{k + 1} (2 π)} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(x + i n)}^{k + 1}} + \frac{1}{{(x - i n)}^{k + 1}}

\underset{n = 1}{\sum^{\infty}} \frac{n^{k}}{e^{2 π n} - 1} = \frac{k!}{{(2 π)}^{k + 1}} \underset{x = 1}{\sum^{\infty}} \frac{1}{x^{k + 1}} + \underset{n = 1}{\sum^{\infty}} \underset{x = 1}{\sum^{\infty}} \frac{1}{{(x + i n)}^{k + 1}} + \frac{1}{{(x - i n)}^{k + 1}}

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n^{k}}{e^{2 π n} - 1} = \frac{k!}{{(2 π)}^{k + 1}} ζ (k + 1) (k = 4 m + 1)

\sum_{x ⩾ 1} \sum_{n ⩾ 1} \frac{1}{{(x + i n)}^{k + 1}} + \frac{1}{{(x - i n)}^{k + 1}} = 0

Commented by Dwaipayan Shikari last updated on 10/Jan/21

F o r e v e r y k = 4 m + 1, t h i s i s v a l i d

\underset{n = 1}{\sum^{\infty}} \frac{n^{5}}{e^{2 π n} - 1} = \frac{1}{504} \dots