let S(x)= Σ_(n=1) ^∞ (−1)^(n−1) (x^(2n+1) /(4n^2 −1)) 1) find the radius of convergence 2) calculate the sum S(x).


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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
$S(x)=(1/2)Σ_(n=1) ^∞ (−1)^(n−1) ((1/(2n−1)) −(1/(2n+1)))x^(2n+1) 2S(x)= Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n−1))x^(2n+1) +Σ_(n=1) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) =Σ_(n=2) ^∞ (((−1)^n )/(2n+1))x^(2n+3) +Σ_(n=1) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) =Σ_(n=0) ^∞ (((−1)^n x^(2n+3) )/(2n+1)) −x^3 +(x^5 /3) +Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) −x =(x^2 +1) Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) −x^3 +(x^5 /3) let put w(x)= Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) w^′ (x) =Σ_(n=0) ^∞ (−1)^n x^(2n) =Σ_(n=0) ^∞ (−x^2 )^n = (1/(1+x^2 )) ⇒ w(x)= arctanx +λ but λ =w(0) =0⇒ w(x)= arctanx so S(x)=(1/2){(x^2 +1)arctanx +(x^5 /3) −x^3 } .$
$S (x) = \frac{1}{2} \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) x^{2 n + 1}$ $2 S (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n - 1} x^{2 n + 1} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}$ $= \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 3} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 3}}{2 n + 1} - x^{3} + \frac{x^{5}}{3} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}$ $- x$ $= (x^{2} + 1) \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1} - x^{3} + \frac{x^{5}}{3}$ $l e t p u t w (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{2 n + 1}$ $w^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} = \sum_{n = 0}^{\infty} {(- x^{2})}^{n} = \frac{1}{1 + x^{2}}$ $\Rightarrow w (x) = a r c t a n x + λ b u t λ = w (0) = 0 \Rightarrow$ $w (x) = a r c t a n x s o$ $S (x) = \frac{1}{2} {(x^{2} + 1) a r c t a n x + \frac{x^{5}}{3} - x^{3}} .$
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
$S(x)= (1/2){ (x^2 +1)arctan(x) −x−x^3 +(x^5 /3)}$
$S (x) = \frac{1}{2} {(x^{2} + 1) a r c t a n (x) - x - x^{3} + \frac{x^{5}}{3}}$
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