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?
 A: 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 $r=1/2$.
A: If you write out the first few (non-zero) terms, $ \ \frac{1}{2^1} \ + \ \frac{2}{2^2} \ + \ \frac{3}{2^3} \ + \ \ldots \ $ , another approach suggests itself, which is to use a "stacking" of infinite series (this would be along the lines of Jacob Bernoulli-style [1680s] ) :
$$  \frac{1}{2^1} \ + \ \frac{2}{2^2} \ + \ \frac{3}{2^3} \ + \ \ldots \ \ = $$
$$ \frac{1}{2^1} \ + \ \frac{1}{2^2} \ + \ \frac{1}{2^3} \ + \ \ldots $$
$$  \quad \quad  \quad \quad \ \ \ + \ \frac{1}{2^2} \ + \ \frac{1}{2^3} \ + \ \frac{1}{2^4} \ + \ \ldots $$
$$  \quad \quad  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \  + \ \frac{1}{2^3} \ + \ \frac{1}{2^4} \ + \ \frac{1}{2^5} \ + \ \ldots \ \ = $$
$$ 1 \ + \ \frac{1}{2} \ + \ \frac{1}{4} \ + \ \frac{1}{8} \ + \ \ldots \ = \ \ 2 \ \ .  $$
[summing the sub-series in each row individually]
More modern methods (such as described by Ian) are more elegant, but Bernoulli got a lot of "mileage" out of this approach and extensions of it back then. 
