1-let-p-integr-natural-not-0-calculate-arctan-p-p-1-arctan-p-1-p-2-let-S-n-p-1-n-arctan-1-2p-2-find-lim-n-S-n-

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} + \dots . + 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}) .