# Convergence of the series nx^n

Gold Member

## Homework Statement

By finding a closed formula for the nth partial sum ##s_n##,
show that the series ## s=\sum\limits_{n=1}^{\infty}nx^n## converges to ##\dfrac{x}{(1-x)^2}## when ##|x|<1## and diverges otherwise.

## Homework Equations

Maybe ##s=\sum\limits_{n=0}^{\infty}x^n=\dfrac{1}{1-x}## when ##|x|<1##

## The Attempt at a Solution

Finding the ##s_n##
##s_n + 1=1 + \sum\limits_{k=1}^{n}(\sqrt[k]{k}x)^k = 1 + x + 2x^2 + 3x^3 + \dots +nx^n= \dfrac{1-(\sqrt[n]{n}x)^{n+1}}{1-\sqrt[n]{n}x}##
but I don't know if I can get anywhere from here, tried several ways and had no success.

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FactChecker
Gold Member
If you know the n'th partial sum of the power series,
##\sum_{i=1} ^{i=n-1} x^i = \frac {(1-x^n)}{(1-x)}##
, can you take the derivative of both sides and use that?

Felipe Lincoln
Gold Member
If you know the n'th partial sum of the power series,
##\sum_{i=1} ^{i=n-1} x^i = \frac {(1-x^n)}{(1-x)}##
, can you take the derivative of both sides and use that?
I would never think of that! I'll try

EDIT: I got it, thank you so much!

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