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Question Number 137721 by Ñï= last updated on 05/Apr/21

\underset{n = 1}{\sum^{\infty}} \tan^{- 1} \frac{1}{2 n^{2}} = ?

e x c e p t u s e \tan^{- 1} \frac{1}{2 n^{2}} = \tan^{- 1} \frac{1}{2 n - 1} - \tan^{- 1} \frac{1}{2 n + 1}, a n y o t h e r w a y ?

Answered by TANMAY PANACEA last updated on 05/Apr/21

{t a n}^{- 1} (\frac{2}{4 n^{2}})

= {t a n}^{- 1} (\frac{2}{1 + 4 n^{2} - 1})

= {t a n}^{- 1} (\frac{(2 n + 1) - (2 n - 1)}{1 + (2 n + 1) (2 n - 1)})

= {t a n}^{- 1} (2 n + 1) - {t a n}^{- 1} (2 n - 1)

T_{n} = {t a n}^{- 1} (2 n + 1) - {t a n}^{- 1} (2 n - 1)

T_{1} = {t a n}^{- 1} (3) - {t a n}^{- 1} (1)

T_{2} = {t a n}^{- 1} (5) - {t a n}^{- 1} (3)

T_{3} = {t a n}^{- 1} (7) - {t a n}^{- 2} (5)

\dots

\dots

T_{n} = {t a n}^{- 1} (2 n + 1) - {t a n}^{- 1} (2 n - 1)

a d d t h e m

S_{n} = {t a n}^{- 1} (2 n + 1) - {t a n}^{- 1} (1)

w h e n n \to \infty

{t a n}^{- 1} (2 n + 1) \to {t a n}^{- 1} (\infty) = \frac{π}{2}

s o S_{\infty} = \frac{π}{2} - {t a n}^{- 1} (1)

= \frac{π}{2} - \frac{π}{4} = \frac{π}{4} T a n m a y

Commented by Ñï= last updated on 06/Apr/21

t h a n k s s i r

Commented by TANMAY PANACEA last updated on 06/Apr/21

m o s t w e l c o m e