1+(1/1^2 )+(1/2^2 )+...+(1/((1.2)^2 ))+(1/((2.3)^2 ))+...+(1/((1.2.3)^2 ))+(1/((2.3.4)^2 ))+...+(1/((1.2.3.4)^2 ))+..... Or 1+Σ_(n=1) ^∞ (1/n^2 )+Σ_(n=1) ^∞ (1/((n(n+1))^2 ))+Σ^∞ (1/((n(n+1)(n+2))^2 ))+.....


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Question Number 121454 by Dwaipayan Shikari last updated on 08/Nov/20

$1 + \frac{1}{1^{2}} + \frac{1}{2^{2}} + . . . + \frac{1}{{(1.2)}^{2}} + \frac{1}{{(2.3)}^{2}} + . . . + \frac{1}{{(1.2.3)}^{2}} + \frac{1}{{(2.3.4)}^{2}} + . . . + \frac{1}{{(1.2.3.4)}^{2}} + . . . . .$ $O r$ $1 + \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(n (n + 1))}^{2}} + \sum^{\infty} \frac{1}{{(n (n + 1) (n + 2))}^{2}} + . . . . .$
Commented by Dwaipayan Shikari last updated on 08/Nov/20

$I t h i n k a l l t e r m s s h o u l d b e \underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{2}})$
Commented by mindispower last updated on 09/Nov/20

$\frac{s i n (x)}{x} = \prod_{n ⩾ 1} (1 - \frac{x^{2}}{n^{2} π^{2}})$ $x = i \Rightarrow \frac{s i n (i π)}{i π} = Π (1 + \frac{1}{n^{2} π^{2}}) = \frac{s h (π)}{π}$ $l n (\prod_{n ⩾ 1} (1 + \frac{1}{n^{2}})) = \sum_{n ⩾ 1} l n (1 + \frac{1}{n^{2}}) = Σ . \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{{k n}^{2 k}}$ $= \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} \sum_{n ⩾ 1} \frac{1}{n^{2 k}} = \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} ζ (2 k)$ $w e [g e t n i c e i d e n t i t i e$ $Σ \frac{{(- 1)}^{k} ζ (2 k)}{k} = \frac{s h (π)}{π}$
	Answered by Olaf last updated on 08/Nov/20

	$S = \underset{k = 0}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{\underset{p = n}{\prod^{n + k}} p^{2}}$ $S = \underset{k = 0}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{(n - 1)!^{2}}{(n + k)!^{2}}$ $S = \underset{n = 1}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{(n - 1)!^{2}}{(n + k)!^{2}}$ $S = \underset{n = 1}{\sum^{\infty}} (n - 1)!^{2} \underset{k = 0}{\sum^{\infty}} \frac{1}{(n + k)!^{2}}$ $. . . see Bessel functions$
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