TITLE: Calculate the sum of the infinite series $\sum_{n=0}^{\infty} \frac{n}{4^n}$
QUESTION [8 upvotes]: A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation. 
Wonderful. Thank you for all the prompt responses.  Could anyone suggest an alternate technique that does not involve differentiation?

REPLY [6 votes]: A simple probabilistic approach:
Let $X$ be a geometric random variable with probability of success $p$, so that
$$
{\rm P}(X=n)=(1-p)^{n-1} p, \;\; n=1,2,3,\ldots.
$$
Then the expectation of $X$ is 
$$
{\rm E}(X) = \sum\limits_{n = 1}^\infty  {n{\rm P}(X = n)}  = \sum\limits_{n = 1}^\infty  {n(1 - p)^{n - 1} p}  = \frac{p}{{1 - p}}\sum\limits_{n = 1}^\infty  {n(1 - p)^n } .
$$
Thus
$$
\sum\limits_{n = 1}^\infty  {n(1 - p)^n } = \frac{{1 - p}}{p} {\rm E}(X).
$$
On the other hand, from
$$
{\rm E}(X) = p \cdot 1 + (1-p)(1+{\rm E}(X)),
$$
we get
$$
{\rm E}(X)=\frac{1}{p}.
$$
Finally,
$$
\sum\limits_{n = 1}^\infty  {n(1 - p)^n } = \frac{{1 - p}}{p} {\rm E}(X) = \frac{{1 - p}}{p^2}.
$$
Letting $p=3/4$ gives 
$$
\sum\limits_{n = 1}^\infty  {\frac{n}{{4^n }}}  = \frac{{1/4}}{{9/16}} = \frac{4}{9}.
$$