nice-calculus-Evaluate-n-0-1-n-2n-1-

Question Number 123763 by mnjuly1970 last updated on 28/Nov/20

\dots n i c e c a l c u l u s \dots

E v a l u a t e :: \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = ? ? ?

Answered by Snail last updated on 28/Nov/20

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1} = {t a n}^{- 1} (x)

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = {t a n}^{- 1} (1) = π / 4

Commented by mnjuly1970 last updated on 28/Nov/20

t h a n k y o u s o m u c h

Answered by Dwaipayan Shikari last updated on 28/Nov/20

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots

A n o t h e r w a y

c o t x = \frac{1}{x} - \frac{1}{π - x} + \frac{1}{π + x} - \frac{1}{2 π - x} + \frac{1}{2 π + x} + \dots

x = \frac{π}{4} \Rightarrow 1 = \frac{4}{π} - \frac{4}{3 π} + \frac{4}{5 π} - \frac{4}{7 π} + \dots

\frac{π}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots

Commented by mnjuly1970 last updated on 28/Nov/20

v e r y n i c e b r a v o s i r d w a i p a y a n . .