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Question Number 165493 by SANOGO last updated on 02/Feb/22

$$calcullasommedecetteserieentiere$$$$\underset{n=0}{\sum ^{+oo}}\frac{2n+3}{2n+1}{x}^n$$

Answered by TheSupreme last updated on 02/Feb/22

$$S=\underset{n=0}{\sum ^{\mathrm{\infty }}}(1+\frac{2}{2n+1})x^n$$$$S=\frac{1}{1-x}+\mathrm{\Sigma }\frac{2}{2n+1}{x}^n$$$$x=s^2$$$$S=\frac{1}{1-s^2}+\frac{2}{s}\mathrm{\Sigma }\frac{1}{2n+1}{s}^{2n+1}$$$$\mathrm{\Sigma }\frac{1}{2n+1}{s}^{2n+1}=\mathrm{\Sigma }\int s^{2n}=\int \mathrm{\Sigma }{s}^{2n}=\int \frac{1}{1-s^2}=arctanh\left(s\right)$$$$S=\frac{1}{1-x}+\frac{1}{\sqrt{x}}arctanh\left(\sqrt{x}\right)$$$$\mid x\mid <1$$

Answered by Mathspace last updated on 02/Feb/22

$$S=\sum _{n=0}^{\mathrm{\infty }}{x}^n+2\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+1}$$$$for\mid x\mid <1\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+1}=\frac{1}{\sqrt{x}}\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(\sqrt{x}\right)}^{2n+1}}{2n+1}$$$$=\frac{1}{\sqrt{x}}\phi \left(\sqrt{x}\right)with\phi \left(t\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{t^{2n+1}}{2n+1}$$$${\phi }^{{}^{\prime }}\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{t}^{2n}=\frac{1}{1-t^2}\Rightarrow$$$$\phi \left(t\right)=\int \frac{dt}{1-t^2}+C$$$$=\frac{1}{2}\int (\frac{1}{1-t}+\frac{1}{1+t})dt$$$$=\frac{1}{2}ln\mid \frac{1+t}{1-t}\mid \Rightarrow \phi \left(t\right)=\frac{1}{2}ln\mid \frac{1+t}{1-t}\mid +C$$$$\phi \left(0\right)=0=C\Rightarrow \phi \left(t\right)=\frac{1}{2}ln\mid \frac{1+t}{1-t}\mid \Rightarrow$$$$S=\frac{1}{1-x}+\frac{2}{\sqrt{x}}\phi \left(\sqrt{x}\right)$$$$=\frac{1}{1-x}+\frac{1}{\sqrt{x}}ln\mid \frac{1+\sqrt{x}}{1-\sqrt{x}}\mid$$

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