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How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$

I need to compute the sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$ using power series.

Any hints of how should I do that?

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    @Jozef: Note that in the answers to the question linked by Didier you have $\sum nr^{n}$. Your question is a special case for $r=1/2$. (Other questions linked to that one might have answers interesting for you too.)2011-12-26

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Remember that $\sum\limits_{0}^{\infty}x^n = \frac{1}{1-x}$ and $\frac{d}{dx}\frac{1}{1-x} = \frac{d}{dx}\sum_{0}^{\infty}x^n = \sum_{0}^{\infty}\frac{d}{dx}x^n = \sum_{1}^{\infty}nx^{n-1}$ so $x\frac{d}{dx}\frac{1}{1-x} = \sum_{1}^{\infty}nx^{n}$. Using $x = 1/2$ should give you what you want.

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    @Alex: Now it looks great, thank you.2011-12-26
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Here's a hint. You know that $\displaystyle \frac{1}{1-x}=\sum_{n\geqslant 0}x^n$. What if you differentiated both sides?

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So you want to compute $S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots$ Now consider $ f(x)= \frac{x}{2} + \frac{x^2}{2^2} + \frac{x^3}{2^3} + \cdots = \displaystyle\frac{\frac{x}{2}}{1 - \frac{x}{2}}$ From here evaluate the value of f'(1).