find-radius-and-sum-of-n-0-x-2n-2n-1-2-find-n-0-1-2n-1-9-n-

Question Number 29981 by abdo imad last updated on 14/Feb/18

f i n d r a d i u s a n d s u m o f \sum_{n = 0}^{\infty} \frac{x^{2 n}}{2 n + 1}

2) f i n d \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1) 9^{n}} .

Commented by abdo imad last updated on 15/Feb/18

l e t p u t f o r ∣ x ∣< 1 S (x) = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{2 n + 1} \Rightarrow x S (x) = \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{2 n + 1} l e t d e r i v a t e

S (x) + {x S}^{^{'}} (x) = \sum_{n = 0}^{\infty} x^{2 n} = \frac{1}{1 - x^{2}} \Rightarrow S i s s o l u t i o n o f d . e

{x y}^{^{'}} + y = \frac{1}{1 - x^{2}} h . e \Rightarrow {x y}^{^{'}} + y = 0 \Rightarrow \frac{y^{^{'}}}{y} = - \frac{1}{x} \Rightarrow

l n ∣ y ∣= - l n x + c \Rightarrow y = \frac{k}{x} l e t u s e m . v . c w e h a v e

y^{^{'}} = \frac{k^{^{'}} x - k}{x^{2}} (e) \Rightarrow \frac{k^{^{'}} x - k}{x} + \frac{k}{x} = \frac{1}{1 - x^{2}} \Rightarrow k^{^{'}} = \frac{1}{1 - x^{2}}

k (x) = \int \frac{d x}{1 - x^{2}} + λ = \frac{1}{2} (\int \frac{d x}{1 - x} + \int \frac{d x}{1 + x}) + λ

= \frac{1}{2} l n ∣ \frac{1 + x}{1 - x} ∣ + λ \Rightarrow S (x) = \frac{1}{2 x} l n ∣ \frac{1 + x}{1 - x} ∣ + \frac{λ}{x}

λ = {l i m}_{x \to 0} (x S (x) - \frac{1}{2} l n ∣ \frac{1 + x}{1 - x} ∣) = 0 \Rightarrow

S (x) = \frac{1}{2 x} l n ∣ \frac{1 + x}{1 - x} ∣ w i t h - 1 < x < 1 .

w e h a v e \frac{u_{n + 1}}{u_{n}} = \frac{\frac{1}{2 n + 3}}{\frac{1}{2 n + 1}} = \frac{2 n + 1}{2 n + 3_{n \to + \infty}} \to 1 \Rightarrow R = 1

f o r x = 1 o r x = - 1 t h e s e r i e d i v e r g e s .

2) \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1) 9^{n}} = \sum_{n = 0}^{\infty} \frac{{(\frac{1}{3})}^{2 n}}{2 n + 1} = S (\frac{1}{3})

= \frac{3}{2} l n ∣ \frac{\frac{4}{3}}{\frac{2}{3}} ∣= \frac{3}{2} l n (2) .