find the sum of the infinite series tan^(−1) ((2/n^2 ))


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Question Number 30235 by NECx last updated on 18/Feb/18

$f i n d t h e s u m o f t h e i n f i n i t e$ $s e r i e s$ $\tan^{- 1} (\frac{2}{n^{2}})$
Commented by prof Abdo imad last updated on 18/Feb/18

$f o r n ⩾ 1 \frac{2}{n^{2}} = \frac{n + 1 - (n - 1)}{1 + (n + 1) (n - 1)} l e t p u t n = {t a n u}_{n}$ $\frac{2}{n^{2}} = \frac{{t a n u}_{n + 1} - {t a n}_{n - 1}}{1 + {t a n u}_{n} {t a n u}_{n - 1}} = t a n (u_{n + 1} - u_{n - 1})$ $a r c t a n (\frac{2}{n^{2}}) = u_{n + 1} - u_{n - 1} a n d$ $S_{N} = \sum_{n = 1}^{N} a r c t a n (\frac{2}{n^{2}}) = \sum_{n = 1}^{N} ((u_{n + 1} - u_{n}) + (u_{n} - u_{n - 1}))$ $= \sum_{n = 1}^{N} (u_{n + 1} - u_{n}) + \sum_{n = 1}^{N} (u_{n} - u_{n - 1})$ $= u_{N + 1} - u_{1} + u_{N} - u_{0} = a r c t a n (N + 1) + a r c t a n N - \frac{π}{4}$ ${l i m}_{N \to \infty} S_{N} = \frac{π}{2} + \frac{π}{2} - \frac{π}{4} = \frac{3 π}{4} .$
Commented by abdo imad last updated on 18/Feb/18

$a r c t a n m e a n s {t a n}^{- 1} .$
Commented by NECx last updated on 19/Feb/18

$t h a n k y o u s o m u c h$
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