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Question Number 80786 by M±th+et£s last updated on 06/Feb/20

Commented by mind is power last updated on 06/Feb/20

$i w i l l t r y n o t e a s y!$
Commented by mr W last updated on 06/Feb/20

$i t h i n k {i t}^{'} s z e r o d u e t o s y m m e t r y .$ $S_{k} = \underset{m = 1}{\sum^{k}} \underset{\underset{n \neq m}{n = 1}}{\sum^{k}} \frac{1}{m^{2} - n^{2}} = 0$ $s i n c e f o r a p a i r (m = i a n d n = j) t h e r e$ $i s a c o r r e s p o n d i n g p a i r (m = j a n d$ $n = i) :$ $\frac{1}{i^{2} - j^{2}} + \frac{1}{j^{2} - i^{2}} = 0$ $S = \underset{m = 1}{\sum^{\infty}} \underset{\underset{n \neq m}{n = 1}}{\sum^{\infty}} \frac{1}{m^{2} - n^{2}} = \lim_{k \to \infty} S_{k} = 0$
Commented by mr W last updated on 06/Feb/20

$t h e q u e s t i o n c a n b e m o d i f i e d t o$ $S = \underset{m = 1}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{m^{2} + n^{2}} = ?$
Commented by mathmax by abdo last updated on 06/Feb/20

$t r y t o p r o v e t h a t \sum_{n = 1 n \neq m}^{\infty} \frac{1}{n^{2} - m^{2}} = \frac{3}{4 m^{2}}$
Commented by ~blr237~ last updated on 07/Feb/20

$l e t h (m) = \sum_{n \in N} \frac{1}{m^{2} + n^{2}}$ $l e t s t a t e f (z) = \frac{1}{m^{2} + z^{2}}, u s i n g t h e R e s i d u t h e o r e m$ $h (m) = - Σ R e s (f (z) π c o t (π z), z_{k})$ $h (m) = - (\frac{π c o t (i π m)}{2 i m} + \frac{π c o t (- i π m)}{- 2 i m})$ $h (m) = i π \frac{c o t (i π m)}{m} = \frac{π c o t h (π m)}{m}$ $S o, S = \underset{m = 1}{\sum^{\infty}} (\frac{1}{m^{2}} + \frac{π c o t h (π m)}{2 m})$ $S = ζ (2) + \frac{π}{2} \underset{m = 1}{\sum^{\infty}} \frac{c o t h (π m)}{m}$
	Answered by ~blr237~ last updated on 07/Feb/20

	$W e h a v e \forall m ⩾ 2 \underset{n \underset{n \neq m}{=} 1}{\sum^{\infty}} \frac{1}{m^{2} - n^{2}} = - \frac{3}{4 m^{2}} (l e t p r o v e)$ $f (m) = \underset{n = 1 n \neq m}{\sum^{\infty}} \frac{1}{m^{2} - n^{2}} . k n o w i n g \frac{1}{m^{2} - n^{2}} = - \frac{1}{2 m} (\frac{1}{n - m} - \frac{1}{n + m})$ $f (m) = - \frac{1}{2 m} [\underset{n = 1}{\sum^{m - 1}} (\frac{1}{n - m} - \frac{1}{n + m}) + \underset{n = m + 1}{\sum^{\infty}} (\frac{1}{n - m} - \frac{1}{n + m})]$ $f (m) = \frac{1}{2 m} [\underset{p = m + 1}{\sum^{2 m - 1}} \frac{1}{p} + \underset{p = 1}{\sum^{m - 1}} \frac{1}{p}] - \frac{1}{2 m} [\underset{p = 1}{\sum^{\infty}} (\frac{1}{p} - \frac{1}{p + 2 m})]$ $l e t s h o w t h a t g (m) = \underset{p = 1}{\sum^{\infty}} (\frac{1}{p} - \frac{1}{p + m}) = \underset{p = 1}{\sum^{m}} \frac{1}{p}$ $g (m) = \underset{p = 1}{\sum^{\infty}} (\underset{k = 0}{\sum^{m - 1}} (\frac{1}{p + k} - \frac{1}{p + k + 1}))$ $= \underset{k = 0}{\sum^{m - 1}} (\underset{p = 1}{\sum^{\infty}} (\frac{1}{k + p} - \frac{1}{k + p + 1}))$ $= \underset{k = 0}{\sum^{m - 1}} \frac{1}{k + 1} = \underset{k = 1}{\sum^{m}} \frac{1}{k}$ $S o,$ $f (m) = \frac{1}{2 m} (\underset{p = 1}{\sum^{2 m - 1}} \frac{1}{p} - \frac{1}{m}) - \frac{1}{2 m} \underset{p = 1}{\sum^{2 m}} \frac{1}{p}$ $f (m) = - \frac{1}{2 m^{2}} - \frac{1}{2 m} \times \frac{1}{2 m} = - \frac{3}{4 m^{2}}$ $F i n a l l y S = \underset{m = 1}{\sum^{\infty}} f (m) = - \frac{3}{4} ζ (2) = - \frac{π^{2}}{8}$
	Commented by M±th+et£s last updated on 07/Feb/20

	$i t h i n k t h a t {i t}^{^{'}} s \frac{- π^{2}}{8}$
	Commented by ~blr237~ last updated on 07/Feb/20

	$y e s {i t}^{'} s$
	Commented by M±th+et£s last updated on 07/Feb/20

	$t h a n k y o u s i r b r i l i a n t s o l u t i o n$
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