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1-7-e-2pi-1-2-7-e-4pi-1-3-7-e-6pi-1-

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Question Number 123179 by Dwaipayan Shikari last updated on 23/Nov/20
(1^7 /(e^(2π) −1))+(2^7 /(e^(4π) −1))+(3^7 /(e^(6π) −1))+....
$$\frac{\mathrm{1}^{\mathrm{7}} }{{e}^{\mathrm{2}\pi} −\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{7}} }{{e}^{\mathrm{4}\pi} −\mathrm{1}}+\frac{\mathrm{3}^{\mathrm{7}} }{{e}^{\mathrm{6}\pi} −\mathrm{1}}+…. \\ $$
Commented by Lordose last updated on 24/Nov/20
Patiently waiting for solution
$$\mathrm{Patiently}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{solution} \\ $$
Commented by Dwaipayan Shikari last updated on 24/Nov/20
Abel plana formula Σ_(z≥0) ^∞ f(z)= ∫_0 ^∞ f(z)dz +((f(0))/2)+i∫_0 ^∞ ((f(iz)−f(−iz))/(e^(2πz) −1))dz Σ_(n≥1) ^∞ (z^7 /(e^(2πz) −1))+f(0)=∫_0 ^∞ (z^7 /(e^(2πz) −1))+((f(0))/2)+i∫_0 ^∞ ((((−iz^7 )/(e^(2iπz) −1))−((iz^7 )/(e^(−2iπz) −1)))/(e^(2πz) −1))dz Σ_(z≥1) ^∞ (z^7 /(e^(2πz) −1))=∫_0 ^∞ ((((u/(2π)))^7 )/((e^u −1)))du −i∫((iz^7 )/(e^(2πz) −1))dz−((f(0))/2) = (1/((2π)^8 ))∫_0 ^∞ (u^7 /(e^u −1))du+(1/((2π)^8 ))∫_0 ^∞ (u^7 /(e^u −1))du 2πz =u = 2((ζ(8)Γ(8))/((2π)^8 ))=((7!)/(2^7 π^8 )).(π^8 /(9450))=((7!)/(2^7 .30)).(1/(5.7.9))=((5!)/(25.9.2^7 ))=((24)/(5.3.3.2^7 )) =(2^3 /(5.3.2^7 ))=(1/(5.3.2^4 ))=(1/(240)) ((f(0))/2)=lim_(z→0) (z^7 /(2(e^(2πz) −1)))=(1/2) ((z^7 /(1+2πz−1)))=0
$${Abel}\:{plana}\:{formula}\: \\ $$$$\underset{{z}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}{f}\left({z}\right)=\:\int_{\mathrm{0}} ^{\infty} {f}\left({z}\right){dz}\:+\frac{{f}\left(\mathrm{0}\right)}{\mathrm{2}}+{i}\int_{\mathrm{0}} ^{\infty} \frac{{f}\left({iz}\right)−{f}\left(−{iz}\right)}{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}{dz} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{7}} }{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}+{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{z}^{\mathrm{7}} }{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}+\frac{{f}\left(\mathrm{0}\right)}{\mathrm{2}}+{i}\int_{\mathrm{0}} ^{\infty} \frac{\frac{−{iz}^{\mathrm{7}} }{{e}^{\mathrm{2}{i}\pi{z}} −\mathrm{1}}−\frac{{iz}^{\mathrm{7}} }{{e}^{−\mathrm{2}{i}\pi{z}} −\mathrm{1}}}{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}{dz} \\ $$$$\:\underset{{z}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{7}} }{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}=\int_{\mathrm{0}} ^{\infty} \frac{\left(\frac{{u}}{\mathrm{2}\pi}\right)^{\mathrm{7}} }{\left({e}^{{u}} −\mathrm{1}\right)}{du}\:−{i}\int\frac{{iz}^{\mathrm{7}} }{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}{dz}−\frac{{f}\left(\mathrm{0}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\left(\mathrm{2}\pi\right)^{\mathrm{8}} }\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{7}} }{{e}^{{u}} −\mathrm{1}}{du}+\frac{\mathrm{1}}{\left(\mathrm{2}\pi\right)^{\mathrm{8}} }\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{7}} }{{e}^{{u}} −\mathrm{1}}{du}\:\:\:\:\:\:\mathrm{2}\pi{z}\:={u}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\frac{\zeta\left(\mathrm{8}\right)\Gamma\left(\mathrm{8}\right)}{\left(\mathrm{2}\pi\right)^{\mathrm{8}} }=\frac{\mathrm{7}!}{\mathrm{2}^{\mathrm{7}} \pi^{\mathrm{8}} }.\frac{\pi^{\mathrm{8}} }{\mathrm{9450}}=\frac{\mathrm{7}!}{\mathrm{2}^{\mathrm{7}} .\mathrm{30}}.\frac{\mathrm{1}}{\mathrm{5}.\mathrm{7}.\mathrm{9}}=\frac{\mathrm{5}!}{\mathrm{25}.\mathrm{9}.\mathrm{2}^{\mathrm{7}} }=\frac{\mathrm{24}}{\mathrm{5}.\mathrm{3}.\mathrm{3}.\mathrm{2}^{\mathrm{7}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{5}.\mathrm{3}.\mathrm{2}^{\mathrm{7}} }=\frac{\mathrm{1}}{\mathrm{5}.\mathrm{3}.\mathrm{2}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{240}} \\ $$$$\frac{{f}\left(\mathrm{0}\right)}{\mathrm{2}}=\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{z}^{\mathrm{7}} }{\mathrm{2}\left({e}^{\mathrm{2}\pi{z}} −\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\:\left(\frac{{z}^{\mathrm{7}} }{\mathrm{1}+\mathrm{2}\pi{z}−\mathrm{1}}\right)=\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 24/Nov/20
I tried but this not the actual answer
$${I}\:{tried}\:{but}\:{this}\:{not}\:{the}\:{actual}\:{answer} \\ $$

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