Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that 
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$ 
I suspect some clever use of the geometric series will do, but I don't know how. 

We can proceed with generating functions. The trick initially seems a little contrived, but it can be made rigorous, and I think that the more you see it the more it makes sense. The first implication comes from differentiation.
$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n \implies \frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}n \cdot x^{n-1}$$
So we can multiply by $x$ to get
$$\frac{x}{(1-x)^2}=\sum_{n=0}^{\infty} n \cdot x^n=\sum_{n=1}^{\infty}n \cdot x^n.$$
Substituting $x=1/3$, we obtain
$\frac{1/3}{(2/3)^2}=3/4$
