Prove-n-1-1-sinh-pin-2-1-6-1-2pi-

Question Number 143098 by mnjuly1970 last updated on 10/Jun/21

\dots . . P r o v e \dots .

\underset{n = 1}{\sum^{\infty}} {(\frac{1}{s i n h (π n)})}^{2} = \frac{1}{6} - \frac{1}{2 π} \dots

\dots \dots

Answered by Dwaipayan Shikari last updated on 10/Jun/21

4 \underset{n = 1}{\sum^{\infty}} \frac{e^{2 π n}}{{(e^{2 π n} - 1)}^{2}}

\frac{\sin h (π x)}{π x} = \underset{n = 1}{\prod^{\infty}} (1 + \frac{x^{2}}{n^{2}})

\Rightarrow c o t h (π x) - \frac{1}{π x} = \underset{n = 1}{\sum^{\infty}} \frac{\frac{2 x}{n^{2}}}{1 + \frac{x^{2}}{n^{2}}}

\Rightarrow \frac{e^{2 π x} + 1}{e^{2 π x} - 1} - \frac{1}{π x} = 2 \underset{n = 1}{\sum^{\infty}} \frac{x}{n^{2} + x^{2}}

\Rightarrow 1 + \frac{2}{e^{2 π x} - 1} - \frac{1}{π x} = 2 \underset{n = 1}{\sum^{\infty}} \frac{x}{(n^{2} + x^{2})}

\frac{- 4 π e^{2 π x}}{{(e^{2 π x} - 1)}^{2}} + \frac{1}{π x^{2}} = - 2 \underset{n = 1}{\sum^{\infty}} \frac{n^{2} + x^{2} - 2 x}{{(n^{2} + x^{2})}^{2}}

\Rightarrow \underset{x = 1}{\sum^{\infty}} \frac{e^{2 π x}}{{(e^{2 π x} - 1)}^{2}} = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{1}{π^{2} x^{2}} + \frac{1}{2} \underset{x = 1}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{(n^{2} + x^{2})} - \frac{2 x}{{(n^{2} + x^{2})}^{2}}

= \frac{1}{4} (. \frac{π^{2}}{π^{2} 6}) + \frac{1}{2} Φ = \frac{1}{24} + Φ \dots

Commented by Dwaipayan Shikari last updated on 10/Jun/21

S o r r y s i r . I h a v e t r i e d

Commented by mnjuly1970 last updated on 10/Jun/21

g r a t e f u l f o r y o u r e f f o r t m r p a y a n . .