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Question Number 127889 by A8;15: last updated on 02/Jan/21

Answered by mindispower last updated on 02/Jan/21

$$=\sum _{n⩾1}\frac{1}{5^{n}n}$$$$\frac{1}{1-x}=\sum _{n⩾0}{x}^n$$$$\Rightarrow {\int }_0^t\frac{1}{1-x}dx\widehat{=}\sum _{n⩾1}\frac{t^n}{n},,t=\frac{1}{5}\Rightarrow$$$${\int }_0^{\frac{1}{5}}\frac{dx}{1-x}=\sum _{n⩾1}\frac{1}{n{5}^n}=-ln(1-\frac{1}{5})=ln\left(\frac{5}{4}\right)$$

Commented by A8;15: last updated on 03/Jan/21

$$\mathrm{Thanks}\mathrm{sir}.\mathbf{Happy}\mathbf{New}\mathbf{Year}$$

Commented by mindispower last updated on 03/Jan/21

$$withepleasur$$$$happynewyear$$

Answered by mr W last updated on 02/Jan/21

$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$$$\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...$$$$\Rightarrow \ln \frac{1}{1-x}=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$$$$withx=\frac{1}{5}$$$$\Rightarrow \frac{1}{5}+\frac{1}{5^2\times 2}+\frac{1}{5^3\times 3}+\frac{1}{5^4\times 4}+...=\ln \frac{5}{4}$$

Commented by A8;15: last updated on 03/Jan/21

$$\mathrm{Thanks}\mathrm{Mr}.\mathrm{W}.\mathbf{Happy}\mathbf{New}\mathbf{Year}$$

Commented by mr W last updated on 03/Jan/21

$$happynewyeartoo!$$

Answered by Dwaipayan Shikari last updated on 03/Jan/21

$$\frac{1}{5}+\frac{1}{5^2.2}+\frac{1}{5^3.3}+...$$$$=-log(1-\frac{1}{5})=log\left(\frac{5}{4}\right)$$$$Generally\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{nk}^n}=log\left(\frac{k}{k-1}\right)$$

Commented by A8;15: last updated on 03/Jan/21

$$\mathrm{Thanks}\mathrm{sir}.\mathbf{Happy}\mathbf{New}\mathbf{Year}$$

Commented by Dwaipayan Shikari last updated on 03/Jan/21

$$Haveagreatyear!$$

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