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Question Number 34697 by abdo imad last updated on 10/May/18

$$letf\left(x\right)=\frac{ln(1+x)}{1+x}$$$$developpfatintegrserie$$

Commented by math khazana by abdo last updated on 11/May/18

$$wehave{ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{n}with\mid x\mid <1$$$$\Rightarrow ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{x}^{n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^n$$$$f\left(x\right)=\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n\right)\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{x}^n\right)$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{x}^n+\left(\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n{x}^n\right)\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{x}^n\right)$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{x}^n+\sum _{n=1}^{\mathrm{\infty }}{c}_n{x}^{n}with$$$$c_n=\sum _{i+j=n}{a}_i{b}_j=\sum _{i=1}^n{a}_i{b}_{n-i}$$$$=\sum _{i=1}^{n-1}{(-1)}^i.\frac{{(-1)}^{n-i}}{n-i}=\sum _{i=1}^{n-1}\frac{{(-1)}^n}{n-i}so$$$$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}{x}^n+\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{i=1}^{n-1}\frac{{(-1)}^n}{n-i}\right)x^n.$$$$$$

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