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Question Number 166291 by SANOGO last updated on 17/Feb/22

$$calculerlasomme$$$$\underset{n=1}{\sum ^{+oo}}\frac{1}{n(n+2)}{x}^n$$

Answered by TheSupreme last updated on 18/Feb/22

$$\mathrm{\Sigma }=a_n{x}^n$$$$\rho =li\underset{n}{m}\frac{a_n}{a_{n+1}}=1$$$$\mid x\mid ⩽1$$$$$$$$\mathrm{\Sigma }=\frac{1}{2}\sum _n(\frac{1}{n}-\frac{1}{n+2})x^n$$$$\mathrm{\Sigma }=\frac{1}{2}\mathrm{\Sigma }\frac{x^n}{n}-\frac{1}{2}\mathrm{\Sigma }\frac{x^n}{n+2}$$$$\mathrm{\Sigma }\frac{x^n}{n}=\mathrm{\Sigma }\int x^{n-1}dx=\int \mathrm{\Sigma }{x}^{n-1}=\int \frac{1}{1-x}dx=-ln\mid 1-x\mid$$$$\mathrm{\Sigma }\frac{x^n}{n+2}=\frac{1}{x^2}\mathrm{\Sigma }\frac{x^{n+2}}{n+2}=\mathrm{\Sigma }\int x^{n+1}dx=\int \mathrm{\Sigma }{x}^{n+1}dx=\int (\frac{1}{1-x}-1-x)dx$$$$-ln\mid 1-x\mid -x-\frac{x^2}{2}$$$$\mathrm{\Sigma }\frac{x^n}{n(n+2)}=\frac{1}{2}(\frac{x^2}{2}+x)$$

Commented by SANOGO last updated on 18/Feb/22

$$mercibien$$

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