advanced-mathematcs-prove-that-n-1-1-n-1-n-2-csch-pi-1-2-

Question Number 130889 by mnjuly1970 last updated on 30/Jan/21

\dots a d v a n c e d m a t h e m a t c s \dots

p r o v e t h a t ::

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

Answered by mindispower last updated on 30/Jan/21

l e t f (z) = \frac{π}{s i n (π z) (1 + z^{2})}, p o l = Z \cup {i, - i}

\int_{C} f (z) d z = 0

R e s (f, k) = \frac{1}{{(- 1)}^{k} (1 + k^{2})}, \forall k \in Z

R e s (f, i) = \frac{π}{2 i s i n (i π)} = \frac{π}{- 2 s h (π)}

R e s (f, - i) = \frac{π}{- s h (π)}

\Rightarrow \sum_{n \in Z} \frac{{(- 1)}^{n}}{1 + n^{2}} - \frac{π}{s h (π)} = 0

\Rightarrow 2 \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{1 + n^{2}} + 1 - \frac{π}{s h (π)} = 0

\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{1 + n^{2}} = \frac{π}{2 s h (π)} - \frac{1}{2} = \frac{π s c c h (π) - 1}{2}

Commented by mnjuly1970 last updated on 30/Jan/21

t a y e b a l l a h m r p o w e r . .

t h a n k y o u \dots

Commented by mindispower last updated on 30/Jan/21

w i t h e p l e a s u r g o d b l e s s y o u