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. 

Take a geometric series with term $r$ and take derivative with respect to $r$ both sides:
$\frac{d}{dr}\sum_{n=0}^{\infty}r^n=\frac{d}{dr}\frac{1}{1-r}.$
Derivate,
$\sum_{n=0}^{\infty}nr^{n-1}=\frac{1}{(1-r)^2}.$
At the left hand side, for $n=0$ the term is zero, so we can start the sum at $n=1$, and also multiply $r$ both sides to balance the exponent:
$r\sum_{n=1}^{\infty}nr^{n-1}=r\frac{1}{(1-r)^2}.$
Now you have
$\sum_{n=1}^{\infty}nr^{n}=\frac{r}{(1-r)^2}.$
For $r=\frac{1}{3}$,
$\sum_{n=1}^{\infty}n\left(\frac{1}{3}\right)^{n}=\frac{\frac{1}{3}}{\left(1-\frac{1}{3}\right)^2}.$
Operate,
$\sum_{n=1}^{\infty}n\frac{1}{3^n}=\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{3}{4}.$
