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Question Number 114996 by mnjuly1970 last updated on 22/Sep/20

                ...nice   mathematics...       prove that:::                        i::  Σ_(n=1) ^∞ (1/(sinh^2 (πn))) =(1/6) −(1/(2π))    ✓                       ii:: Σ_(n=1) ^∞ (n/(e^(2πn) −1))=(1/(24)) −(1/(8π))  ✓✓                      iii::Σ_(n=1) ^∞ (1/( nsinh(πn))) =(π/(12))−((ln(2))/4) ✓✓✓                  ....    M..n..july..1970 ....

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:\:{mathematics}...\: \\ $$$$\:\:\:\:{prove}\:{that}::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{sinh}^{\mathrm{2}} \left(\pi{n}\right)}\:=\frac{\mathrm{1}}{\mathrm{6}}\:−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\:\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{24}}\:−\frac{\mathrm{1}}{\mathrm{8}\pi}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{iii}::\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:{nsinh}\left(\pi{n}\right)}\:=\frac{\pi}{\mathrm{12}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{4}}\:\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\:\:\:\:\mathscr{M}..{n}..{july}..\mathrm{1970}\:.... \\ $$$$ \\ $$

Commented by maths mind last updated on 24/Sep/20

i will poste all my worck later   just ii) i found2((1/(24))−(1/(8π)))  ...

$${i}\:{will}\:{poste}\:{all}\:{my}\:{worck}\:{later}\: \\ $$$$\left.{just}\:{ii}\right)\:{i}\:{found}\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{8}\pi}\right)\:\:... \\ $$

Answered by Olaf last updated on 23/Sep/20

i::  I_n  = ∫_n ^(n+1) (dx/(sinh^2 (πx)))  I_n  = ∫_n ^(n+1) (coth^2 (πx)−1)dx  I_n  = [(−(1/π)coth(πx)]_n ^(n+1) = (1/π)[coth(πn)−coth(π(n+1))]  I_n   = (1/π)[((cosh(πn))/(sinh(πn)))−((cosh(π(n+1)))/(sinh(π(n+1))))]  I_n   =(1/π) ((sinh(π(n+1))cosh(πn)−cosh(π(n+1)+cosh(πn))/(sinh(πn)sinh(π(n+1))))  I_n   = (1/π)[((sinh(π(n+1)−πn])/(sinh(πn)sinh(π(n+1))))]  I_n   = (1/π)[((sinh(π))/(sinh(πn)sinh(π(n+1))))]  (π/(sinh(π)))I_n  = (1/(sinh(πn)sinh(π(n+1))))  (π/(sinh(π)))I_(n+1)  ≤ (1/(sinh^2 (π(n+1)))) ≤ (π/(sinh(π)))I_n   (π/(sinh(π)))Σ_(n=1) ^∞ I_(n+1)  ≤ Σ_(n=1) ^∞ (1/(sinh^2 (π(n+1)))) ≤Σ_(n=1) ^∞  (π/(sinh(π)))I_n   (π/(sinh(π)))Σ_(n=2) ^∞ I_n  ≤ Σ_(n=1) ^∞ (1/(sinh^2 (πn)))−(1/(sinh^2 (π))) ≤Σ_(n=1) ^∞  (π/(sinh(π)))I_n   (π/(sinh(π)))[−(1/π)coth(πx)]_2 ^∞  ≤ Σ_(n=1) ^∞ (1/(sinh^2 (πn)))−(1/(sinh^2 (π))) ≤ (π/(sinh(π)))[−(1/π)coth(πx)]_1 ^∞   (1/(sinh(π)))[coth(2π)−1] ≤ Σ_(n=1) ^∞ (1/(sinh^2 (πn)))−(1/(sinh^2 (π))) ≤ (1/(sinh(π)))[coth(π)−1]  (1/(sinh(π)))[coth(2π)−1+(1/(sinh(π)))] ≤ Σ_(n=1) ^∞ (1/(sinh^2 (πn))) ≤ (1/(sinh(π)))[coth(π)−1+(1/(sinh(π)))]_1 ^∞   ((coth(2π)sinh(π)−sinh(π)+1)/(sinh^2 (π)))  ≤ Σ_(n=1) ^∞ (1/(sinh^2 (πn))) ≤ [((cosh(π)−sinh(π)+1)/(sinh^2 (π)))]  I tried but may be it is not the good way.

$${i}:: \\ $$$$\mathrm{I}_{{n}} \:=\:\int_{{n}} ^{{n}+\mathrm{1}} \frac{{dx}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi{x}\right)} \\ $$$$\mathrm{I}_{{n}} \:=\:\int_{{n}} ^{{n}+\mathrm{1}} \left(\mathrm{coth}^{\mathrm{2}} \left(\pi{x}\right)−\mathrm{1}\right){dx} \\ $$$$\mathrm{I}_{{n}} \:=\:\left[\left(−\frac{\mathrm{1}}{\pi}\mathrm{coth}\left(\pi{x}\right)\right]_{{n}} ^{{n}+\mathrm{1}} =\:\frac{\mathrm{1}}{\pi}\left[\mathrm{coth}\left(\pi{n}\right)−\mathrm{coth}\left(\pi\left({n}+\mathrm{1}\right)\right)\right]\right. \\ $$$$\mathrm{I}_{{n}} \:\:=\:\frac{\mathrm{1}}{\pi}\left[\frac{\mathrm{cosh}\left(\pi{n}\right)}{\mathrm{sinh}\left(\pi{n}\right)}−\frac{\mathrm{cosh}\left(\pi\left({n}+\mathrm{1}\right)\right)}{\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)\right)}\right] \\ $$$$\mathrm{I}_{{n}} \:\:=\frac{\mathrm{1}}{\pi}\:\frac{\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)\right)\mathrm{cosh}\left(\pi{n}\right)−\mathrm{cosh}\left(\pi\left({n}+\mathrm{1}\right)+\mathrm{cosh}\left(\pi{n}\right)\right.}{\mathrm{sinh}\left(\pi{n}\right)\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)\right)} \\ $$$$\mathrm{I}_{{n}} \:\:=\:\frac{\mathrm{1}}{\pi}\left[\frac{\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)−\pi{n}\right]}{\mathrm{sinh}\left(\pi{n}\right)\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)\right)}\right] \\ $$$$\mathrm{I}_{{n}} \:\:=\:\frac{\mathrm{1}}{\pi}\left[\frac{\mathrm{sinh}\left(\pi\right)}{\mathrm{sinh}\left(\pi{n}\right)\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)\right)}\right] \\ $$$$\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\mathrm{I}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi{n}\right)\mathrm{sinh}\left(\pi\left({n}+\mathrm{1}\right)\right)} \\ $$$$\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\mathrm{I}_{{n}+\mathrm{1}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\left({n}+\mathrm{1}\right)\right)}\:\leqslant\:\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\mathrm{I}_{{n}} \\ $$$$\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{I}_{{n}+\mathrm{1}} \:\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\left({n}+\mathrm{1}\right)\right)}\:\leqslant\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\mathrm{I}_{{n}} \\ $$$$\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\mathrm{I}_{{n}} \:\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi{n}\right)}−\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\right)}\:\leqslant\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\mathrm{I}_{{n}} \\ $$$$\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\left[−\frac{\mathrm{1}}{\pi}\mathrm{coth}\left(\pi{x}\right)\right]_{\mathrm{2}} ^{\infty} \:\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi{n}\right)}−\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\right)}\:\leqslant\:\frac{\pi}{\mathrm{sinh}\left(\pi\right)}\left[−\frac{\mathrm{1}}{\pi}\mathrm{coth}\left(\pi{x}\right)\right]_{\mathrm{1}} ^{\infty} \\ $$$$\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi\right)}\left[\mathrm{coth}\left(\mathrm{2}\pi\right)−\mathrm{1}\right]\:\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi{n}\right)}−\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\right)}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi\right)}\left[\mathrm{coth}\left(\pi\right)−\mathrm{1}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi\right)}\left[\mathrm{coth}\left(\mathrm{2}\pi\right)−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi\right)}\right]\:\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi{n}\right)}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi\right)}\left[\mathrm{coth}\left(\pi\right)−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sinh}\left(\pi\right)}\right]_{\mathrm{1}} ^{\infty} \\ $$$$\frac{\mathrm{coth}\left(\mathrm{2}\pi\right)\mathrm{sinh}\left(\pi\right)−\mathrm{sinh}\left(\pi\right)+\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\right)}\:\:\leqslant\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi{n}\right)}\:\leqslant\:\left[\frac{\mathrm{cosh}\left(\pi\right)−\mathrm{sinh}\left(\pi\right)+\mathrm{1}}{\mathrm{sinh}^{\mathrm{2}} \left(\pi\right)}\right] \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{but}\:\mathrm{may}\:\mathrm{be}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{the}\:\mathrm{good}\:\mathrm{way}. \\ $$

Answered by maths mind last updated on 24/Sep/20

ii)  Σ(n/(e^(2πn) −1))  let f(z)=(z/(e^(2πz) −1))    holomorphic cunction over C−{iZ}  it has rvable singularity at origine  lim_(z→0)   (z/(e^(2πz) −1))=(1/(2π))  we can use  Σ_(n≥0) f(n)=(1/(2π))+Σ_(n≥1) f(n)  Σ_(n≥0) f(n)=∫_0 ^∞ f(x)dx+((f(0))/2)+i∫_0 ^∞ ((f(it)−f(−it))/(e^(2πt) −1))dt Abel −plana formula  f(it)−f(−it)  =((it)/(e^(2iπt) −1))+((it)/(e^(−2iπt) −1)) =−it⇒i∫_0 ^∞ ((f(it)−f(−it))/(e^(2πt) −1))dt=∫_0 ^∞ (x/(e^(2πx) −1))dx  Σ_(n≥0) f(n)_(=S) =(1/(2π))+Σ_(n≥1) f(n)=(1/(2π))+Σ_(n≥1) f(n)=(1/(4π))+2∫_0 ^∞ ((xdx)/(e^(2πx) −1))  =2∫_0 ^∞ ((xdx)/(e^(2πx) −1)),2πx=t⇒dx=(dt/(2π))⇒  S=(1/(4π))−(1/(2π))+(1/(2π^2 ))∫_0 ^∞ (t/(e^t −1))dt=−(1/(4π))+(1/(2π^2 ))ζ(2)Γ(2)=−(1/(4π))+(1/(12)).  Σ(n/(e^(2πn) −1))=(1/(12))−(1/(4π))

$$\left.{ii}\right) \\ $$$$\Sigma\frac{{n}}{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}} \\ $$$${let}\:{f}\left({z}\right)=\frac{{z}}{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}\:\:\:\:{holomorphic}\:{cunction}\:{over}\:{C}−\left\{{iZ}\right\} \\ $$$${it}\:{has}\:{rvable}\:{singularity}\:{at}\:{origine} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{z}}{{e}^{\mathrm{2}\pi{z}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}\pi} \\ $$$${we}\:{can}\:{use} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}+\underset{{n}\geqslant\mathrm{1}} {\sum}{f}\left({n}\right) \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{f}\left({n}\right)=\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right){dx}+\frac{{f}\left(\mathrm{0}\right)}{\mathrm{2}}+{i}\int_{\mathrm{0}} ^{\infty} \frac{{f}\left({it}\right)−{f}\left(−{it}\right)}{{e}^{\mathrm{2}\pi{t}} −\mathrm{1}}{dt}\:{Abel}\:−{plana}\:{formula} \\ $$$${f}\left({it}\right)−{f}\left(−{it}\right) \\ $$$$=\frac{{it}}{{e}^{\mathrm{2}{i}\pi{t}} −\mathrm{1}}+\frac{{it}}{{e}^{−\mathrm{2}{i}\pi{t}} −\mathrm{1}}\:=−{it}\Rightarrow{i}\int_{\mathrm{0}} ^{\infty} \frac{{f}\left({it}\right)−{f}\left(−{it}\right)}{{e}^{\mathrm{2}\pi{t}} −\mathrm{1}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{x}}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{f}\left({n}\underset{={S}} {\right)}=\frac{\mathrm{1}}{\mathrm{2}\pi}+\underset{{n}\geqslant\mathrm{1}} {\sum}{f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}+\underset{{n}\geqslant\mathrm{1}} {\sum}{f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{4}\pi}+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{xdx}}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{xdx}}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}},\mathrm{2}\pi{x}={t}\Rightarrow{dx}=\frac{{dt}}{\mathrm{2}\pi}\Rightarrow \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{4}\pi}−\frac{\mathrm{1}}{\mathrm{2}\pi}+\frac{\mathrm{1}}{\mathrm{2}\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{t}}{{e}^{{t}} −\mathrm{1}}{dt}=−\frac{\mathrm{1}}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{2}\pi^{\mathrm{2}} }\zeta\left(\mathrm{2}\right)\Gamma\left(\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{12}}. \\ $$$$\Sigma\frac{{n}}{{e}^{\mathrm{2}\pi{n}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}\pi} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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