find-the-value-of-n-0-1-n-2n-3-

Question Number 33886 by math khazana by abdo last updated on 26/Apr/18

f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 3} .

Commented by math khazana by abdo last updated on 29/Apr/18

c h a n g e m e n t o f i n d i c e n = p - 1 g i v e

S = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{2 (p - 1) + 3} = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{2 p + 1}

= - \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p}}{2 p + 1} = - \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{2 p + 1} + 1

w (x) = \sum_{p = o}^{\infty} \frac{{(- 1)}^{p}}{2 p + 1} x^{2 p + 1} w i t h ∣ x ∣< 1

w^{^{'}} (x) = \sum_{p = 0}^{\infty} {(- 1)}^{p} x^{2 p} = \sum_{p = 0}^{\infty} {(- x^{2})}^{p} = \frac{1}{1 + x^{2}}

\Rightarrow w (x) = \int_{0}^{x} \frac{d t}{1 + t^{2}} + λ b u t λ = w (0) = 0 \Rightarrow

w (x) = a r c t a n x

S = 1 - w (1) = 1 - \frac{π}{4}

★ S = 1 - \frac{π}{4} ★