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Question Number 29979 by abdo imad last updated on 14/Feb/18

$$findtheradiusofS\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{3n+2}}{3n+2}$$$$2)findthevalueof\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+2)3^n}.$$

Commented by abdo imad last updated on 17/Feb/18

$$1)forx=0S\left(x\right)=0forx\ne owehave$$$$\mid \frac{u_{n+1}\left(x\right)}{u_n\left(x\right)}\mid =\mid \frac{\frac{x^{3n+5}}{3n+5}}{\frac{x^{3n+2}}{3n+2}}\mid =\frac{3n+2}{3n+5}\mid x{\mid }_{n\to \mathrm{\infty }}^3\to \mid x{\mid }^3\Rightarrow if\mid x\mid <1$$$$theserieconvergeif\mid x\mid ⩾1theseriedivergeifx=-1$$$$S(-1)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n+2}converge\left(alternateserie\right)soR=1$$$$2)for\mid x\mid <1andduetouniformconvergencewehave$$$$S^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^{3n+1}=x\sum _{n=0}^{\mathrm{\infty }}{\left(x^3\right)}^n=x\frac{1}{1-x^3}\Rightarrow$$$$S\left(x\right)=\int \frac{xdx}{1-x^3}+\lambda letdecompose$$$$F\left(x\right)=\frac{x}{1-x^3}=\frac{x}{(1-x)(x^2+x+1)}=\frac{a}{1-x}+\frac{bx+c}{x^2+x+1}$$$$a={lim}_{x\to 1}(1-x)F\left(x\right)=\frac{1}{3}so$$$$F\left(x\right)=\frac{1}{3(1-x)}+\frac{bx+c}{x^2+x+1}$$$${lim}_{x\to \mathrm{\infty }}xF\left(x\right)=0=\frac{-1}{3}+b\Rightarrow b=\frac{1}{3}\Rightarrow$$$$F\left(x\right)=\frac{1}{3(1-x)}+\frac{\frac{1}{3}x+c}{x^2+x+1}$$$$F\left(0\right)=0=\frac{1}{3}+c\Rightarrow c=-\frac{1}{3}and$$$$F\left(x\right)=\frac{1}{3(1-x)}+\frac{1}{3}\frac{x-1}{x^2+x+1}\Rightarrow$$$$\int \frac{x}{1-x^3}dx=\frac{1}{3}(\int \frac{dx}{1-x}+\int \frac{x-1}{x^2+x+1}dx)$$$$3\int \frac{x}{1-x^3}dx=\int \frac{dx}{1-x}+\frac{1}{2}\int \frac{2x+1-3}{x^2+x+1}dx$$$$=\frac{1}{2}ln(x^2+x+1)-ln\mid 1-x\mid -\frac{3}{2}\int \frac{dx}{x^2+x+1}but$$$$\int \frac{dx}{x^2+x+1}=\int \frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}{=}_{x+\frac{1}{2}=\frac{\sqrt{3}}{2}t}\int \frac{1}{\frac{3}{4}(t^2+1)}\frac{\sqrt{3}}{2}dt$$$$=\frac{4}{3}\frac{\sqrt{3}}{2}\int \frac{dt}{1+t^2}=\frac{2}{\sqrt{3}}artant=\frac{2}{\sqrt{3}}arctan\left(\frac{2}{\sqrt{3}}(x+\frac{1}{2})\right)$$$$3\int \frac{x}{1-x^2}dx=ln\left(\frac{\sqrt{x^2+x+1}}{\mid 1-x\mid }\right)-\sqrt{3}arctan\left(\frac{1}{\sqrt{3}}(2x+1)\right)$$$$\int \frac{xdx}{1-x^2}=\frac{1}{3}ln\left(\frac{\sqrt{x^2+x+1}}{\mid 1-x\mid }\right)-\frac{\sqrt{3}}{3}arctan\left(\frac{1}{\sqrt{3}}(2x+1)\right)$$$$S\left(x\right)=\frac{1}{3}ln\left(\frac{\sqrt{x^2+x+1}}{\mid 1-x\mid }\right)-\frac{\sqrt{3}}{3}arctan\left(\frac{1}{\sqrt{3}}(2x+1)\right)+\lambda$$$$=A\left(x\right)+\lambda \Rightarrow \lambda ={lim}_{\to 0}S\left(x\right)-A\left(x\right)=-A\left(0\right)$$$$=\frac{\sqrt{3}}{3}arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\sqrt{3}}{3}\frac{\pi }{6}=\frac{\pi \sqrt{3}}{18}andS\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{3n+2}}{3n+2}$$$$wehave\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+2)3^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(\frac{1}{3}\right)}^n}{3n+2}$$$$={\left(3_{\sqrt{3}}\right)}^2\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(\frac{1}{3_{\sqrt{3}}}\right)}^{3n+2}}{3n+2}={\left(3_{\sqrt{3}}\right)}^{2}S\left({\left(3_{\sqrt{3}}\right)}^{-1}\right)thesumSis$$$$known.$$

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