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

$$findradiusandsumof\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{2n+1}$$$$2)find\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)9^n}.$$

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

$$letputfor\mid x\mid <1S\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{2n+1}\Rightarrow xS\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n+1}}{2n+1}letderivate$$$$S\left(x\right)+{xS}^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^{2n}=\frac{1}{1-x^2}\Rightarrow Sissolutionofd.e$$$${xy}^{{}^{\prime }}+y=\frac{1}{1-x^2}h.e\Rightarrow {xy}^{{}^{\prime }}+y=0\Rightarrow \frac{y^{{}^{\prime }}}{y}=-\frac{1}{x}\Rightarrow$$$$ln\mid y\mid =-lnx+c\Rightarrow y=\frac{k}{x}letusem.v.cwehave$$$$y^{{}^{\prime }}=\frac{k^{{}^{\prime }}x-k}{x^2}\left(e\right)\Rightarrow \frac{k^{{}^{\prime }}x-k}{x}+\frac{k}{x}=\frac{1}{1-x^2}\Rightarrow k^{{}^{\prime }}=\frac{1}{1-x^2}$$$$k\left(x\right)=\int \frac{dx}{1-x^2}+\lambda =\frac{1}{2}(\int \frac{dx}{1-x}+\int \frac{dx}{1+x})+\lambda$$$$=\frac{1}{2}ln\mid \frac{1+x}{1-x}\mid +\lambda \Rightarrow S\left(x\right)=\frac{1}{2x}ln\mid \frac{1+x}{1-x}\mid +\frac{\lambda }{x}$$$$\lambda ={lim}_{x\to 0}(xS\left(x\right)-\frac{1}{2}ln\mid \frac{1+x}{1-x}\mid )=0\Rightarrow$$$$S\left(x\right)=\frac{1}{2x}ln\mid \frac{1+x}{1-x}\mid with-1<x<1.$$$$wehave\frac{u_{n+1}}{u_n}=\frac{\frac{1}{2n+3}}{\frac{1}{2n+1}}=\frac{2n+1}{2n+3_{n\to +\mathrm{\infty }}}\to 1\Rightarrow R=1$$$$forx=1orx=-1theseriediverges.$$$$2)\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)9^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(\frac{1}{3}\right)}^{2n}}{2n+1}=S\left(\frac{1}{3}\right)$$$$=\frac{3}{2}ln\mid \frac{\frac{4}{3}}{\frac{2}{3}}\mid =\frac{3}{2}ln\left(2\right).$$

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