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Question Number 123179 by Dwaipayan Shikari last updated on 23/Nov/20

$$\frac{1^7}{e^{2\pi }-1}+\frac{2^7}{e^{4\pi }-1}+\frac{3^7}{e^{6\pi }-1}+....$$

Commented by Lordose last updated on 24/Nov/20

$$\mathrm{Patiently}\mathrm{waiting}\mathrm{for}\mathrm{solution}$$

Commented by Dwaipayan Shikari last updated on 24/Nov/20

$$Abelplanaformula$$$$\underset{z⩾0}{\sum ^{\mathrm{\infty }}}f\left(z\right)={\int }_0^{\mathrm{\infty }}f\left(z\right)dz+\frac{f\left(0\right)}{2}+i{\int }_0^{\mathrm{\infty }}\frac{f\left(iz\right)-f(-iz)}{e^{2\pi z}-1}dz$$$$\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{z^7}{e^{2\pi z}-1}+f\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{z^7}{e^{2\pi z}-1}+\frac{f\left(0\right)}{2}+i{\int }_0^{\mathrm{\infty }}\frac{\frac{-{iz}^7}{e^{2i\pi z}-1}-\frac{{iz}^7}{e^{-2i\pi z}-1}}{e^{2\pi z}-1}dz$$$$\underset{z⩾1}{\sum ^{\mathrm{\infty }}}\frac{z^7}{e^{2\pi z}-1}={\int }_0^{\mathrm{\infty }}\frac{{\left(\frac{u}{2\pi }\right)}^7}{(e^u-1)}du-i\int \frac{{iz}^7}{e^{2\pi z}-1}dz-\frac{f\left(0\right)}{2}$$$$=\frac{1}{{\left(2\pi \right)}^8}{\int }_0^{\mathrm{\infty }}\frac{u^7}{e^u-1}du+\frac{1}{{\left(2\pi \right)}^8}{\int }_0^{\mathrm{\infty }}\frac{u^7}{e^u-1}du2\pi z=u$$$$=2\frac{\zeta \left(8\right)\mathrm{\Gamma }\left(8\right)}{{\left(2\pi \right)}^8}=\frac{7!}{2^7{\pi }^8}.\frac{{\pi }^8}{9450}=\frac{7!}{2^7.30}.\frac{1}{5.7.9}=\frac{5!}{25.9.2^7}=\frac{24}{5.3.3.2^7}$$$$=\frac{2^3}{5.3.2^7}=\frac{1}{5.3.2^4}=\frac{1}{240}$$$$\frac{f\left(0\right)}{2}=\underset{z\to 0}{\lim }\frac{z^7}{2(e^{2\pi z}-1)}=\frac{1}{2}\left(\frac{z^7}{1+2\pi z-1}\right)=0$$

Commented by Dwaipayan Shikari last updated on 24/Nov/20

$$Itriedbutthisnottheactualanswer$$

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