find Σ_(n=1) ^(+∞) arctan( ((ln(1+(1/(n+1))) −ln(1+(1/n)))/(1+(1 +(1/n))(1+(1/(n+1))))))


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Question Number 33718 by prof Abdo imad last updated on 22/Apr/18

$f i n d \sum_{n = 1}^{+ \infty} a r c t a n (\frac{l n (1 + \frac{1}{n + 1}) - l n (1 + \frac{1}{n})}{1 + (1 + \frac{1}{n}) (1 + \frac{1}{n + 1)}})$
Commented by prof Abdo imad last updated on 26/Apr/18

$l e t p u t S_{n} = \sum_{n = 1}^{N} a r c t a n (\frac{l n (1 + \frac{1}{n + 1}) - l n (1 + \frac{1}{n})}{1 + l n (1 + \frac{1}{n}) l n (1 + \frac{1}{n + 1})})$ $l e t p u t l n (1 + \frac{1}{n}) = t a n (u_{n}) \Rightarrow u_{n} = a r c t a n (l n (1 + \frac{1}{n}))$ $S_{n} = \sum_{k = 1}^{n} a r c t a n (\frac{t a n (u_{n + 1}) - t a n (u_{n})}{1 + t a n (u_{n}) t a n (u_{n + 1})})$ $= \sum_{k = 1}^{N} a r c t a n (t a n (u_{n + 1} - u_{n}))$ $= \sum_{k = 1}^{N} (u_{n + 1} - u_{n}) = u_{N + 1} - u_{1} = a r c t a n (l n (1 + \frac{1}{N + 1})$ $- a r c t a n (l n (2)) a n d$ ${l i m}_{n \to + \infty} S_{n} = a r c t a n (l n (1)) - a r c t a n (l n (2))$ ${l i m}_{n \to + \infty} S_{n} = - a r c t a n (l n (2)) .$
Commented by prof Abdo imad last updated on 26/Apr/18

$t h e Q . i s f i n d \sum_{n = 1}^{+ \infty} a r c t a n (\frac{l n (1 + \frac{1}{n + 1}) - l n (1 + \frac{1}{n})}{1 + l n (1 + \frac{1}{n}) l n (1 + \frac{1}{n + 1})})$
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