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Question Number 221306 by Gbenga last updated on 30/May/25
Σ_(n=1) ^∞ ((csch^2 (πn))/n^2 )
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{csch}^{\mathrm{2}} \left(\pi{n}\right)}{{n}^{\mathrm{2}} } \\ $$
Answered by SdC355 last updated on 30/May/25
Σ_(l=1) ^∞ ((csch^2 (πl))/l^2 )=Σ_(l=1) ^∞ (4/(l^2 (e^(πl) −e^(−πl) )^2 ))=0.00750124....
$$\underset{{l}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{csch}^{\mathrm{2}} \left(\pi{l}\right)}{{l}^{\mathrm{2}} }=\underset{{l}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{4}}{{l}^{\mathrm{2}} \left({e}^{\pi{l}} −{e}^{−\pi{l}} \right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{00750124}…. \\ $$
Commented by Gbenga last updated on 30/May/25
is there a way to find the closed form
$${is}\:{there}\:{a}\:{way}\:{to}\:{find}\:{the}\:{closed}\:{form} \\ $$
Commented by SdC355 last updated on 30/May/25
nope... can′t find Closed form of Σ ((csch^2 (kπ))/k^2 ) ∫_1 ^( ∞) (((csch(πz))/z))^2 dz ≤Σ ((csch^2 (πk))/k^2 )≤∫_0 ^( ∞) (((csch(πz))/z))^2 dz L_1 ≤Σ ((csch^2 (kπ))/k^2 )≤L_2 and L_1 ,L_2 are bounded So Σ ((csch^2 (kπ))/k^2 ) is convergence
$$\mathrm{nope}… \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{Closed}\:\mathrm{form}\:\mathrm{of}\:\:\Sigma\:\frac{\mathrm{csch}^{\mathrm{2}} \left({k}\pi\right)}{{k}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{1}} ^{\:\infty} \:\left(\frac{\mathrm{csch}\left(\pi{z}\right)}{{z}}\right)^{\mathrm{2}} \mathrm{d}{z}\:\leq\Sigma\:\:\frac{\mathrm{csch}^{\mathrm{2}} \left(\pi{k}\right)}{{k}^{\mathrm{2}} }\leq\int_{\mathrm{0}} ^{\:\infty} \:\:\left(\frac{\mathrm{csch}\left(\pi{z}\right)}{{z}}\right)^{\mathrm{2}} \mathrm{d}{z}\: \\ $$$${L}_{\mathrm{1}} \leq\Sigma\:\:\frac{\mathrm{csch}^{\mathrm{2}} \left({k}\pi\right)}{{k}^{\mathrm{2}} }\leq{L}_{\mathrm{2}} \\ $$$$\mathrm{and}\:{L}_{\mathrm{1}} ,{L}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{bounded}\: \\ $$$$\mathrm{So}\:\Sigma\:\:\frac{\mathrm{csch}^{\mathrm{2}} \left({k}\pi\right)}{{k}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{convergence} \\ $$
Commented by Gbenga last updated on 30/May/25
Answered by MrGaster last updated on 31/May/25
ζ(2)=Σ_(m=1) ^∞ (1/m^2 )=(π^2 /6) Σ_(m=1) ^∞ (1/(n^2 sinh^2 (πn)))=Σ_(m=1) ^∞ (4/(m^2 (e^(mπ) −e^(−mπ) )^2 )) =4Σ_(m=1) ^∞ (e^(−2mπ) /(m^2 (1−e^(−2mπ) )^2 )) =4Σ_(m=1) ^∞ (e^(−2mπ) /n^2 )Σ_(k=0) ^∞ (k+1)e^(−2kmπ) =4Σ_(k=0) ^∞ (k+1)Σ_(m=1) ^∞ (e^(−2(k+1)mπ) /m^2 ) =4Σ_(k=0) ^∞ kΣ_(m=1) ^∞ (e^(−2kmπ) /m^2 ) =4Σ_(k=1) ^∞ k(Li_2 (e^(−2kπ) )−(π^2 /6)+π log(1−e^(−2kπ) )) Li_2 (e^(−2kπ) )=Σ_(m=1) ^∞ (e^(−2kmπ) /m^2 ) Σ_(k=1) ^∞ kLi_2 (e^(−2kπ) )=(G/π)−(π/(72)) Σ_(k=1) ^∞ k log(1−e^(−2kπ) )=(π/(12))+(1/4)log(2π) Σ_(k=1) ^∞ k(−(π^2 /6))=−(π^2 /6)ζ(−1)=(π^2 /(72)) 4((G/π)−(π/(72))−(π^3 /(72))+π(−(π/(12))+(1/4)log(2π))) =((4G)/π)−(π/(18))−(π^3 /(18))−(π^2 /3)+π log(2π) =((2G)/3)−((11π^2 )/(180))
$$\zeta\left(\mathrm{2}\right)=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{m}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \mathrm{sinh}^{\mathrm{2}} \left(\pi{n}\right)}=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}}{{m}^{\mathrm{2}} \left({e}^{{m}\pi} −{e}^{−{m}\pi} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{−\mathrm{2}{m}\pi} }{{m}^{\mathrm{2}} \left(\mathrm{1}−{e}^{−\mathrm{2}{m}\pi} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{−\mathrm{2}{m}\pi} }{{n}^{\mathrm{2}} }\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left({k}+\mathrm{1}\right){e}^{−\mathrm{2}{km}\pi} \\ $$$$=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left({k}+\mathrm{1}\right)\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{−\mathrm{2}\left({k}+\mathrm{1}\right){m}\pi} }{{m}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{k}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{−\mathrm{2}{km}\pi} }{{m}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{k}\left(\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{k}\pi} \right)−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\pi\:\mathrm{log}\left(\mathrm{1}−{e}^{−\mathrm{2}{k}\pi} \right)\right) \\ $$$$\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{k}\pi} \right)=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{−\mathrm{2}{km}\pi} }{{m}^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{k}\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{k}\pi} \right)=\frac{{G}}{\pi}−\frac{\pi}{\mathrm{72}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{k}\:\mathrm{log}\left(\mathrm{1}−{e}^{−\mathrm{2}{k}\pi} \right)=\frac{\pi}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\left(\mathrm{2}\pi\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{k}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\zeta\left(−\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{72}} \\ $$$$\mathrm{4}\left(\frac{{G}}{\pi}−\frac{\pi}{\mathrm{72}}−\frac{\pi^{\mathrm{3}} }{\mathrm{72}}+\pi\left(−\frac{\pi}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\left(\mathrm{2}\pi\right)\right)\right) \\ $$$$=\frac{\mathrm{4}{G}}{\pi}−\frac{\pi}{\mathrm{18}}−\frac{\pi^{\mathrm{3}} }{\mathrm{18}}−\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\pi\:\mathrm{log}\left(\mathrm{2}\pi\right) \\ $$$$=\frac{\mathrm{2}{G}}{\mathrm{3}}−\frac{\mathrm{11}\pi^{\mathrm{2}} }{\mathrm{180}} \\ $$

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