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Question Number 49961 by maxmathsup by imad last updated on 12/Dec/18

$$findthesequence\left(a_n\right)wichverify$$$$\left(\sum _{n=1}^{\mathrm{\infty }}{x}^n\right)\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-x)}^n}{n+1}\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n}alsofindtheradiusofthisserie.$$

Commented by Abdo msup. last updated on 13/Dec/18

$$let{\alpha }_n=1\mathrm{\forall }n\in Nand{\beta }_n=\frac{{(-1)}^n}{n+1}so$$$$\left(\sum _{n=1}^{\mathrm{\infty }}{x}^n\right).\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{x}^n\right)=\left(\sum _{n=0}^{\mathrm{\infty }}{\alpha }_n{x}^n\right).\left(\sum _{n=0}^{\mathrm{\infty }}{\beta }_n{x}^n\right)$$$$\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n/a_n=\sum _{i+j=n}{\alpha }_i{\beta }_j=\sum _{i=0}^n\frac{{(-1)}^{n-i}}{n-i+1}$$$$\Rightarrow a_n=\sum _{p=0}^n\frac{{(-1)}^p}{p+1}=\sum _{p=1}^{n+1}\frac{{(-1)}^{p-1}}{p}$$$$wehaveRadius\left(\mathrm{\Sigma }{x}^n\right)=1andRadiud\left(\mathrm{\Sigma }\frac{{(-x)}^n}{n+1}\right)=1\Rightarrow$$$$Re>adius\left(\mathrm{\Sigma }{a}_n{x}^n\right)⩽1.$$

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