let-give-S-n-p-1-p-n-arctan-1-2p-2-find-lim-n-gt-S-n-

Question Number 27189 by abdo imad last updated on 02/Jan/18

l e t g i v e S_{n} = \sum_{p = 1}^{p = n} a r c t a n (\frac{1}{2 p^{2}}) f i n d {l i m}_{n - >\propto} S_{n} .

Answered by prakash jain last updated on 03/Jan/18

\tan^{- 1} \frac{1}{2 p^{2}} = \tan^{- 1} \frac{(2 p + 1) - (2 p - 1)}{1 + (2 p + 1) (2 p + 1)}

= \tan^{- 1} (2 p + 1) - \tan^{- 1} (2 p - 1)

p : 1 \to \tan^{- 1} 3 - \tan^{- 1} 1

p : 2 \to \tan^{- 1} 5 - \tan^{- 1} 3

⋮

p : n - 1 \to \tan^{- 1} (2 n - 1) - \tan^{- 1} (2 n - 3)

p : n \to \tan^{- 1} (2 n + 1) - \tan^{- 1} (2 n - 3 > 1)

\underset{p = 1}{\sum^{n}} \tan^{- 1} \frac{1}{2 p^{2}} = \tan^{- 1} (2 n + 1) - \tan^{- 1} 1

\lim_{n \to \infty} \underset{p = 1}{\sum^{n}} \tan^{- 1} \frac{1}{2 p^{2}} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}