let-S-x-n-1-1-n-1-x-2n-1-4n-2-1-1-find-the-radius-of-convergence-2-calculate-the-sum-S-x-

Question Number 34309 by prof Abdo imad last updated on 03/May/18

l e t S (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{2 n + 1}}{4 n^{2} - 1}

1) f i n d t h e r a d i u s o f c o n v e r g e n c e

2) c a l c u l a t e t h e s u m S (x) .

Commented by math khazana by abdo last updated on 07/May/18

S (x) = \frac{1}{2} \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) x^{2 n + 1}

2 S (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n - 1} x^{2 n + 1} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}

= \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 3} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 3}}{2 n + 1} - x^{3} + \frac{x^{5}}{3} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}

- x

= (x^{2} + 1) \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1} - x^{3} + \frac{x^{5}}{3}

l e t p u t w (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}

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

\Rightarrow w (x) = a r c t a n x + λ b u t λ = w (0) = 0 \Rightarrow

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

S (x) = \frac{1}{2} {(x^{2} + 1) a r c t a n x + \frac{x^{5}}{3} - x^{3}} .

Commented by math khazana by abdo last updated on 07/May/18

t h e r a d i u s o f c o n v e r g e n c e i s R = 1 .

Commented by math khazana by abdo last updated on 08/May/18

S (x) = \frac{1}{2} {(x^{2} + 1) a r c t a n (x) - x - x^{3} + \frac{x^{5}}{3}}