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Question Number 125659 by Mammadli last updated on 12/Dec/20
(1/(2020^1 ))+(2/(2020^2 ))+(3/(2020^3 ))+...+(n/(2020^n ))=?
$$\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{\boldsymbol{{n}}}{\mathrm{2020}^{\boldsymbol{{n}}} }=? \\ $$
Answered by Dwaipayan Shikari last updated on 12/Dec/20
S=(1/(2020))+(2/(2020^2 ))+...+(n/(2020^n )) (S/(2020))= (1/(2020^2 ))+(2/(2020^3 ))+...+((n−1)/(2020^n ))+(n/(2020^(n+1) )) S−(S/(2020))=(1/(2020))+(1/(2020^2 ))+(1/(2020^3 ))+..(1/(2020^n ))−(n/(2020^(n+1) )) ((2019S)/(2020))=((1−(2020)^(−n) )/(2019))−(n/(2020^(n+1) )) S=((2020−(2020)^(1−n) )/(2019^2 ))−(n/(2019.2020^n ))
$${S}=\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+…+\frac{{n}}{\mathrm{2020}^{{n}} } \\ $$$$\frac{{S}}{\mathrm{2020}}=\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{{n}−\mathrm{1}}{\mathrm{2020}^{{n}} }+\frac{{n}}{\mathrm{2020}^{{n}+\mathrm{1}} } \\ $$$${S}−\frac{{S}}{\mathrm{2020}}=\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{3}} }+..\frac{\mathrm{1}}{\mathrm{2020}^{{n}} }−\frac{{n}}{\mathrm{2020}^{{n}+\mathrm{1}} } \\ $$$$\frac{\mathrm{2019}{S}}{\mathrm{2020}}=\frac{\mathrm{1}−\left(\mathrm{2020}\right)^{−{n}} }{\mathrm{2019}}−\frac{{n}}{\mathrm{2020}^{{n}+\mathrm{1}} } \\ $$$${S}=\frac{\mathrm{2020}−\left(\mathrm{2020}\right)^{\mathrm{1}−{n}} }{\mathrm{2019}^{\mathrm{2}} }−\frac{{n}}{\mathrm{2019}.\mathrm{2020}^{{n}} } \\ $$

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