# Using power series, compute: $\sum\limits_{n=1}^{\infty}\frac{n}{2^n}$ [duplicate]

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?

• @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.) – Martin Sleziak Dec 26 '11 at 9:05

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.
• The fundamental fact about power series is that for $|x|<1$, $\sum_{0}^{\infty}x^n = \frac{1}{1-x}$. Since these two expressions are equal, their derivatives must be equal. – Alex Becker Dec 26 '11 at 9:39
• For interchanging sum and derivative: I think we can either use uniform convergence, since power series converges uniformly for $|x|<r$, where $r$ is the radius of convergence. Or - perhaps easier approach - we can consider the series as formal power series. – Martin Sleziak Dec 26 '11 at 9:52
Here's a hint. You know that $\displaystyle \frac{1}{1-x}=\sum_{n\geqslant 0}x^n$. What if you differentiated both sides?
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 $f'(1)$.