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 ....

Commented bymaths mind last updated on 24/Sep/20

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

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.

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))