nice-calculus-prove-that-n-1-1-n-2-pi-2-1-1-e-2-1-

Question Number 138163 by mnjuly1970 last updated on 10/Apr/21

\dots \dots . . n i c e \dots \dots . \dots . c a l c u l u s \dots . .

p r o v e t h a t ::

Ψ = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n^{2} π^{2} + 1}) \overset{? ? ?}{=} \frac{1}{e^{2} - 1}

\dots \dots \dots \dots .

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

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

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

= \frac{1}{2} (\frac{e^{2} + 1}{e^{2} - 1} - 1) = \frac{1}{e^{2} - 1} = \frac{\frac{1}{e}}{e - \frac{1}{e}}

Commented by mnjuly1970 last updated on 10/Apr/21

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