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Question Number 27098 by abdo imad last updated on 02/Jan/18

$$letgiveS\left(x\right)=\sum _{n=1}^{\propto }\frac{x^n}{n}andW\left(x\right)=\sum _{n=1}^{\propto }\frac{{(-1)}^n{x}^n}{n^2}$$ $$calculateS\left(x\right).W\left(x\right).inthatweknow/x/<1.$$

Commented byprakash jain last updated on 02/Jan/18

$$\ln (1-x)=-x-\frac{x^2}{2}-...\left(A\right)$$ $$=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n}$$ $$\Rightarrow S\left(x\right)=-\ln (1-x)$$ $$\ln (1+x)dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{x}^n}{n}$$ $${\int }_0^x\ln (1+x)dx={\int }_0^x\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{x}^n}{n}$$ $$(x+1)\ln (1+x)-x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{x}^{n+1}}{n^2}$$ $$(x+1)\ln (1+x)-x=-x\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^n}{n^2}$$ $$W\left(x\right)=\frac{(x+1)\ln (1+x)-x}{x}$$ $$\mathrm{Given}\mid x\mid <1\mathrm{both}S\left(x\right)\mathrm{and}W\left(x\right)$$ $$\mathrm{converge}.$$

Commented byabdo imad last updated on 04/Jan/18

$$letcalculateS\left(x\right)W\left(x\right)informofseriesweput{a}_n=\frac{1}{n}$$ $$and{b}_n=\frac{{(-1)}^n}{n^2}$$ $$S\left(x\right)W\left(x\right)=\sum _{n=1}^{\propto }{c}_n{x}^{n}with{c}_n=\sum _{i+j=n}{a}_i{b}_j$$ $$c_n=\sum _{i=1}^{n-1}{a}_i{b}_{n-i}$$ $$S\left(x\right)W\left(x\right)=\sum _{n=1}^{\propto }\left(\sum _{i=1}^{n-1}\frac{1}{i}\frac{{(-1)}^{n-i}}{{(n-i)}^2}\right)x^n.$$ $$$$

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