... advanced mathematcs ... prove that:: Σ_(n=1) ^∞ (((−1)^n )/(1+n^2 )) =((csch(π)−1)/2)


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

$. . . a d v a n c e d m a t h e m a t c s . . .$ $p r o v e t h a t ::$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{1 + n^{2}} = \frac{c s c h (π) - 1}{2}$
	Answered by mindispower last updated on 30/Jan/21
	$let f(z)=(π/(sin(πz)(1+z^2 ))),pol =Z∪{i,−i} ∫_C f(z)dz=0 Res(f,k)=(1/((−1)^k (1+k^2 ))),∀k∈Z Res(f,i)=(π/(2isin(iπ)))=(π/(−2sh(π))) Res(f,−i)=(π/(−sh(π))) ⇒Σ_(n∈Z) (((−1)^n )/(1+n^2 ))−(π/(sh(π)))=0 ⇒2Σ_(n≥1) (((−1)^n )/(1+n^2 ))+1−(π/(sh(π)))=0 Σ_(n≥1) (((−1)^n )/(1+n^2 ))=(π/(2sh(π)))−(1/2)=((πscch(π)−1)/2)$
	$l e t f (z) = \frac{π}{s i n (π z) (1 + z^{2})}, p o l = Z \cup {i, - i}$ $\int_{C} f (z) d z = 0$ $R e s (f, k) = \frac{1}{{(- 1)}^{k} (1 + k^{2})}, \forall k \in Z$ $R e s (f, i) = \frac{π}{2 i s i n (i π)} = \frac{π}{- 2 s h (π)}$ $R e s (f, - i) = \frac{π}{- s h (π)}$ $\Rightarrow \sum_{n \in Z} \frac{{(- 1)}^{n}}{1 + n^{2}} - \frac{π}{s h (π)} = 0$ $\Rightarrow 2 \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{1 + n^{2}} + 1 - \frac{π}{s h (π)} = 0$ $\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{1 + n^{2}} = \frac{π}{2 s h (π)} - \frac{1}{2} = \frac{π s c c h (π) - 1}{2}$
	Commented by mnjuly1970 last updated on 30/Jan/21

	$t a y e b a l l a h m r p o w e r . .$ $t h a n k y o u . . .$
	Commented by mindispower last updated on 30/Jan/21

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