1) let p integr natural not 0 calculate arctan((p/(p+1)))−arctan(((p−1)/p)) 2)let S_n =Σ_(p=1) ^n arctan((1/(2p^2 ))) find lim_(n→+∞) S


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Question Number 51985 by maxmathsup by imad last updated on 01/Jan/19

$1) l e t p i n t e g r n a t u r a l n o t 0 c a l c u l a t e a r c t a n (\frac{p}{p + 1}) - a r c t a n (\frac{p - 1}{p})$ $2) l e t S_{n} = \sum_{p = 1}^{n} a r c t a n (\frac{1}{2 p^{2}}) f i n d {l i m}_{n \to + \infty} S_{n}$
Commented by Abdo msup. last updated on 19/Jan/19

$w e h a v e t a n (a r c t a n (\frac{p}{p + 1}) - a r c t a n (\frac{p - 1}{p}))$ $= \frac{\frac{p}{p + 1} - \frac{p - 1}{p}}{1 + \frac{p}{p + 1} \frac{p - 1}{p}} = \frac{p^{2} - p^{2} + 1}{p^{2} + p + p^{2} - p} = \frac{1}{2 p^{2}} .$ $2) w e h a v e a r c t a n (\frac{1}{2 p^{2}}) = a r c t a n (\frac{p}{p + 1}) - a r c t a n (\frac{p - 1}{p})$ $w e {h a v e S}_{n} = \sum_{p = 1}^{n} a r c t a n (\frac{1}{2 p^{2}}) \Rightarrow$ $S_{n} = \sum_{p = 1}^{n} (a r c t a n (\frac{p}{p + 1}) - a r c t a n (\frac{p - 1}{p}))$ $= \sum_{p = 1}^{n} (U_{p} - U_{p - 1}) (U_{p} = a r c t a n (\frac{p}{p + 1}))$ $= U_{1} - U_{0} + U_{2} - U_{1} + . . . . + U_{n} - U_{n - 1}$ $= U_{n} - U_{1} = a r c t a n (\frac{n}{n + 1}) - a r c t a n (\frac{1}{2}) \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = \frac{π}{4} - a r c t a n (\frac{1}{2}) .$
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