show-that-n-1-arctan-2-n-2-3pi-4-

Question Number 98060 by M±th+et+s last updated on 11/Jun/20

s h o w t h a t

\underset{n = 1}{\sum^{\infty}} a r c t a n (\frac{2}{n^{2}}) = \frac{3 π}{4}

Answered by smridha last updated on 11/Jun/20

i s i t r i g h t a n s ? ? ? I t h i n k {i t}^{'} s j u s t

\frac{π}{4} \dots a n d a l s o I p r o v e t h i s . .

Commented by M±th+et+s last updated on 11/Jun/20

Commented by smridha last updated on 11/Jun/20

\underset{n = 1}{\sum^{\infty}} \tan^{- 1} [\frac{(n + 1) - (n - 1)}{1 + (n + 1) (n - 1)}]

= \underset{n = 0}{\sum^{\infty}} [\tan^{- 1} (n + 1) - \tan^{- 1} (n - 1)]

n o w p a r t i a l s u m o f s e q u e n c e . .

S_{1} = \tan^{- 1} (2) - \tan^{- 1} (0)

S_{2} = \tan^{- 1} (2) + \tan^{- 1} (3) - \tan^{- 1} (1)

S_{3} = \tan^{- 1} (3) + \tan^{- 1} (4) - \tan^{- 1} (1)

\dots \dots \dots \dots \dots \dots

\dots \dots \dots \dots \dots \dots

S_{n} = \tan^{- 1} (n + 1) + \tan^{- 1} (n - 1) - \frac{π}{4}

n o w,

\lim_{n \to \infty} S_{n} = \frac{π}{2} + \frac{π}{2} - \frac{π}{4} = \frac{3 π}{4}

s o r r y f o r m i s s j u d g e m e n t a t f i r s t

{i t}^{'} s p e r f e c t . .

o k k t h a n k s

Commented by M±th+et+s last updated on 11/Jun/20

i n t e l l i g n t .

t h a n k y o u s i r