sequences and series - Sum $sum_{x=0}^{infty} frac{x}{2^x}$

Calculate $\sum\limits_{x=0}^{\infty} \dfrac{x}{2^x}$

So, this series converges by ratio test. How do I find the sum? Any hints?

Answer

As a first step, let us prove that

$$f(r) := \sum_{n=0}^\infty r^n = \frac{1}{1-r}$$

if $r \in (-1,1)$. This is the geometric series. If you haven't seen this proven before, here's a proof. Define

$$S_N = \sum_{n=0}^N r^n.$$

Then

$$r S_N = \sum_{n=0}^N r^{n+1} = \sum_{n=1}^{N+1} r^n = S_N - 1 + r^{N+1}.$$

Solve this equation for $S_N$, obtaining

$$S_N = \frac{1-r^{N+1}}{1-r}$$

and send $N \to \infty$ to conclude.

The sum above converges absolutely, so we can differentiate term by term. Doing so we get

$$f'(r) = \sum_{n=0}^\infty n r^{n-1} = \frac{1}{(1-r)^2}.$$

(Precisely speaking, the sum in the middle is ill-defined at $r=0$, in that it has the form $0/0$. However, $f'(0)=1$ still holds. This doesn't matter for this problem, but it should be noted regardless.) Now multiply by $r$ to change it into your form:

$$\sum_{n=0}^\infty n r^n = \frac{r}{(1-r)^2}.$$

Now substitute $r=1/2$.