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

Commented by LUFFY last updated on 10/Jan/21

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

Commented by LUFFY last updated on 10/Jan/21

$$\mathit{s}\mathit{e}\mathit{n}\mathit{d}\mathit{m}\mathit{e}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}$$

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

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^k}{e^{2\pi n}-1}=\frac{k!}{{\left(2\pi \right)}^{k+1}}.\zeta (k+1)(k=4m+1)$$$$k=13Itis\frac{13!}{{\left(2\pi \right)}^{14}}.\frac{{\left(2\pi \right)}^{14}{B}_{14}}{2\left(14\right)!}=\frac{B_{14}}{28}=\frac{1}{24}$$$${\mathbf{B}}_n=BernoulliNumber$$

Commented by LUFFY last updated on 10/Jan/21

$$\mathrm{check}\mathrm{again}$$

Commented by LUFFY last updated on 10/Jan/21

$$\mathrm{can}\mathrm{you}\mathrm{send}\mathrm{full}\mathrm{solution}$$

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

$$\frac{1}{e^{2\pi }-1}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{e}^{-2\pi n}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+x^2}=\frac{\pi }{2x}coth\left(\pi x\right)-\frac{1}{2x^2}$$$$\frac{1}{x}+2x\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+x^2}=\pi coth\left(\pi x\right)\Rightarrow \frac{1}{2x}+x\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+x^2}=\frac{\pi }{2}+\frac{\pi }{e^{2\pi x}-1}$$$$\frac{1}{e^{2\pi x}-1}=\frac{x}{\pi }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+x^2}+\frac{1}{2\pi x}-\frac{1}{2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{e}^{-2\pi nx}=\frac{1}{2\pi x}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{x+in}+\frac{1}{x-in}+\frac{1}{2\pi x}-\frac{1}{2}$$$$differentiatingkthtime(k=4m+1)$$$${\left(2\pi \right)}^k\underset{x=1}{\sum ^{\mathrm{\infty }}}{n}^k{e}^{-2\pi nx}=\frac{k!}{{\left(x\right)}^{k+1}\left(2\pi \right)}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(x+in)}^{k+1}}+\frac{1}{{(x-in)}^{k+1}}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^k}{e^{2\pi n}-1}=\frac{k!}{{\left(2\pi \right)}^{k+1}}\underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{1}{x^{k+1}}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{x=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(x+in)}^{k+1}}+\frac{1}{{(x-in)}^{k+1}}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^k}{e^{2\pi n}-1}=\frac{k!}{{\left(2\pi \right)}^{k+1}}\zeta (k+1)(k=4m+1)$$$$\sum _{x⩾1}\sum _{n⩾1}\frac{1}{{(x+in)}^{k+1}}+\frac{1}{{(x-in)}^{k+1}}=0$$

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

$$Foreveryk=4m+1,thisisvalid$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^5}{e^{2\pi n}-1}=\frac{1}{504}...$$

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