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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}} + \dots + \frac{1}{{(1.2)}^{2}} + \frac{1}{{(2.3)}^{2}} + \dots + \frac{1}{{(1.2.3)}^{2}} + \frac{1}{{(2.3.4)}^{2}} + \dots + \frac{1}{{(1.2.3.4)}^{2}} + \dots . .

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}} + \dots . .

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}}

\dots see Bessel functions

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