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Question Number 110920 by Dwaipayan Shikari last updated on 31/Aug/20

Commented by Dwaipayan Shikari last updated on 31/Aug/20

$$Ihavefoundthiswhileexperimenting$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n{2}^n}=\frac{1}{1.2}+\frac{1}{2.2^2}+....=-log(1-\frac{1}{2})=log\left(2\right)$$$$....$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n.k^n}=log\left(\frac{k}{k-1}\right)$$

Answered by mathmax by abdo last updated on 01/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\mathrm{with}\mid \mathrm{x}\mid <1\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}-1}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}$$$$=\frac{1}{1-\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=-\ln (1-\mathrm{x})+\mathrm{c}\mathrm{we}\mathrm{have}\mathrm{c}=\mathrm{f}\left(0\right)=0\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}=-\ln (1-\mathrm{x})\mathrm{let}\mathrm{change}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}\mathrm{with}\mid \mathrm{x}\mid >1\mathrm{we}\mathrm{get}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{nx}}^{\mathrm{n}}}=-\ln (1-\frac{1}{\mathrm{x}})=-\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)=\ln \left(\frac{\mathrm{x}}{\mathrm{x}-1}\right)$$$$\mathrm{for}\mathrm{x}=\mathrm{k}\mathrm{integr}\mathrm{we}\mathrm{get}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{nk}}^{\mathrm{n}}}=\ln \left(\frac{\mathrm{k}}{\mathrm{k}-1}\right)$$

Commented by Dwaipayan Shikari last updated on 01/Sep/20

$$Thankingyou$$

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