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Question Number 131686 by Dwaipayan Shikari last updated on 07/Feb/21

Σ_(n=1) ^∞ ((coth(nπ))/n^3 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi\right)}{{n}^{\mathrm{3}} } \\ $$

Commented by Dwaipayan Shikari last updated on 07/Feb/21

I have found ((7π^3 )/(180))

$${I}\:{have}\:{found}\:\frac{\mathrm{7}\pi^{\mathrm{3}} }{\mathrm{180}} \\ $$

Commented by Dwaipayan Shikari last updated on 07/Feb/21

Generally  Σ_(n=1) ^∞ ((coth(nπx)+x^2 coth(((nπ)/x)))/n^3 )=(π^3 /(90x))(x^4 +5x^2 +1)  Σ_(n=1) ^∞ ((coth(nπ)+coth(nπ))/n^3 )=((7π^3 )/(90))  Σ_(n=1) ^∞ ((coth(nπ))/n^3 )=((7π^3 )/(180))

$${Generally} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi{x}\right)+{x}^{\mathrm{2}} {coth}\left(\frac{{n}\pi}{{x}}\right)}{{n}^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{90}{x}}\left({x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi\right)+{coth}\left({n}\pi\right)}{{n}^{\mathrm{3}} }=\frac{\mathrm{7}\pi^{\mathrm{3}} }{\mathrm{90}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi\right)}{{n}^{\mathrm{3}} }=\frac{\mathrm{7}\pi^{\mathrm{3}} }{\mathrm{180}} \\ $$

Commented by mnjuly1970 last updated on 07/Feb/21

mr  payan   please  explain a little (of course if  possible)...

$${mr}\:\:{payan}\: \\ $$$${please}\:\:{explain}\:{a}\:{little}\:\left({of}\:{course}\:{if}\right. \\ $$$$\left.{possible}\right)... \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 07/Feb/21

coth(πx)=(1/(πx))+((2x)/π)Σ_(k=1) ^∞ (1/(k^2 +x^2 ))              [from  ((sinh(πx))/(πx)) =Π_(n=1) ^∞ (1+(x^2 /n^2 ))]  coth(πnx)=(1/(πnx))+((2xn)/π)Σ_(k=1) ^∞ (1/(k^2 +n^2 x^2 ))  Σ_(n=1) ^∞ ((coth(πnx)+x^2 coth(((πn)/x)))/n^3 )  =Σ_(n=1) ^∞ (1/(πn^4 x))+((2x)/π)Σ_n Σ_k (1/n^2 ).(1/((k^2 +n^2 x^2 )))+Σ_n (x^3 /(πn^4 ))+((2x^3 )/π)Σ_n Σ_k (1/(n^2 (k^2 x^2 +n^2 )))  =(π^3 /(90x))+((π^3 x^3 )/(90))+((2x)/π)Σ_n Σ_k (1/(n^2 (k^2 +n^2 x^2 )))+((2x^3 )/π)Σ_n Σ_k (1/(n^2 (k^2 x^2 +n^2 )))  =(π^3 /(90x))(x^4 +1)+((2x)/π)(Σ_n Σ_k (k^2 /(n^2 k^2 (k^2 +n^2 x^2 )))+Σ_n Σ_k ((x^2 k^2 )/(n^2 k^2 (k^2 x^2 +n^2 ))))  =(π^3 /(90x))(x^4 +1)+((2x)/π)(Σ_n Σ_k (1/(n^2 k^2 )))  =(π^3 /(90x))(x^4 +1)+((π^3 x)/(18))=(π^3 /(90x))(x^4 +5x^2 +1)

$${coth}\left(\pi{x}\right)=\frac{\mathrm{1}}{\pi{x}}+\frac{\mathrm{2}{x}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{from}\:\:\frac{{sinh}\left(\pi{x}\right)}{\pi{x}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\right] \\ $$$${coth}\left(\pi{nx}\right)=\frac{\mathrm{1}}{\pi{nx}}+\frac{\mathrm{2}{xn}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left(\pi{nx}\right)+{x}^{\mathrm{2}} {coth}\left(\frac{\pi{n}}{{x}}\right)}{{n}^{\mathrm{3}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\pi{n}^{\mathrm{4}} {x}}+\frac{\mathrm{2}{x}}{\pi}\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }.\frac{\mathrm{1}}{\left({k}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}+\underset{{n}} {\sum}\frac{{x}^{\mathrm{3}} }{\pi{n}^{\mathrm{4}} }+\frac{\mathrm{2}{x}^{\mathrm{3}} }{\pi}\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({k}^{\mathrm{2}} {x}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{90}{x}}+\frac{\pi^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{90}}+\frac{\mathrm{2}{x}}{\pi}\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({k}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}+\frac{\mathrm{2}{x}^{\mathrm{3}} }{\pi}\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({k}^{\mathrm{2}} {x}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{90}{x}}\left({x}^{\mathrm{4}} +\mathrm{1}\right)+\frac{\mathrm{2}{x}}{\pi}\left(\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} {k}^{\mathrm{2}} \left({k}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}+\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{{x}^{\mathrm{2}} {k}^{\mathrm{2}} }{{n}^{\mathrm{2}} {k}^{\mathrm{2}} \left({k}^{\mathrm{2}} {x}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)}\right) \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{90}{x}}\left({x}^{\mathrm{4}} +\mathrm{1}\right)+\frac{\mathrm{2}{x}}{\pi}\left(\underset{{n}} {\sum}\underset{{k}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} {k}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{90}{x}}\left({x}^{\mathrm{4}} +\mathrm{1}\right)+\frac{\pi^{\mathrm{3}} {x}}{\mathrm{18}}=\frac{\pi^{\mathrm{3}} }{\mathrm{90}{x}}\left({x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}\right)\: \\ $$

Commented by mnjuly1970 last updated on 07/Feb/21

thank you so much sir payan..   grateful...

$${thank}\:{you}\:{so}\:{much}\:{sir}\:{payan}.. \\ $$$$\:{grateful}... \\ $$

Commented by mnjuly1970 last updated on 07/Feb/21

excellent.bravo...bravo...  very nice..

$${excellent}.{bravo}...{bravo}... \\ $$$${very}\:{nice}.. \\ $$

Answered by rs4089 last updated on 07/Feb/21

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