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

Question

Possible Duplicate:
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?

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?

Answer 1

Here's a hint. You know that $\displaystyle \frac{1}{1-x}=\sum_{n\geqslant 0}x^n$. What if you differentiated both sides?

Answer 2

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.

Answer 3

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 $f'(1)$.