find-the-value-of-n-0-artan-n-1-n-1-n-2-n-

Question Number 33717 by prof Abdo imad last updated on 22/Apr/18

f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} a r t a n (\frac{\sqrt{n + 1} - \sqrt{n}}{1 + \sqrt{n^{2} + n}})

Commented by prof Abdo imad last updated on 24/Apr/18

l e t p u t S_{n} = \sum_{k = 0}^{n} a r c t a n (\frac{\sqrt{k + 1} - \sqrt{k}}{1 + \sqrt{k^{2} + k}}) l e t p u t

\sqrt{k} = t a n (u_{k}) \Leftrightarrow u_{k} = a r c t a n (\sqrt{k}) \Rightarrow

S_{n} = \sum_{k = 0}^{n} a r c t a n (\frac{t a n (u_{k + 1}) - t a n (u_{k})}{1 + t a n (u_{k}) t a n (u_{k + 1})})

= \sum_{k = 0}^{n} a r c t a n (t a n (u_{k + 1} - u_{k}))

= \sum_{k = 0}^{n} (u_{k + 1} - u_{k}) = u_{n + 1} - u_{0} = a r c t a n (\sqrt{n}) \Rightarrow

{l i m}_{n \to + \infty} S_{n} = \frac{π}{2} .

Commented by prof Abdo imad last updated on 24/Apr/18

S_{n} = a r c t a n (\sqrt{n + 1})

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18

\underset{0}{\sum^{\infty}} \tan^{- 1} (\sqrt{n + 1)} - \tan^{- 1} (\sqrt{n)}

T_{n} = \tan^{- 1} (\sqrt{n + 1)} - \tan^{- 1} (\sqrt{n)}

T_{0} = \tan^{- 1} (\sqrt{1)} - \tan^{- 1} (0)

T_{1} = \tan^{- 1} (\sqrt{2)} - \tan^{- 1} (1)

s o S_{n} = T_{0} + T_{1} + \dots . . + T_{n}

= \tan^{- 1} (\sqrt{n + 1}) - \tan^{- 1} (0)

w h e n n t e n d s t o i n f i n i t y t h e s u m i s

\tan^{- 1} (\infty) - \tan^{- 1} (0)

= ∐ / 2 - 0

= ∐ / 2

∐ t h i s s i g n u s e d a s p i e