<s>[INST] 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?
 [/INST] 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. 
</s>