Computing $\sum_{n\geq 0}n\frac{1}{4^n}$ Can I compute the sum
$$
\sum_{n\geq 0}n\frac{1}{4^n}
$$
by use of some trick?
First I thought of a geometrical series?
 A: This is very similar to the geometric series, infact $$\sum_{n=0}^\infty n x^n = x \cdot \frac{d}{dx} \sum_{n=0}^\infty x^n = x \cdot \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$
Plugging in $x=1/4$ will provide you with an answer.
A: Suppose $\sum_{n\ge0}n\dfrac1{4^n}=S$, then $4S=S+\sum_{n\ge0}\dfrac1{4^n}$.
$\Rightarrow3S=\dfrac43$
$\Rightarrow S=\dfrac49$
I think this might be the solution.
A: The sum can be expanded as follows:
\begin{align*}
\begin{array}{l|cccc}
\text{once}&1/4^1&&&\\
\text{twice}&1/4^2&1/4^2&&\\
\text{3 times}&1/4^3&1/4^3&1/4^3&&\\
\vdots\\
\text{$n$ times}&1/4^n&1/4^n&\cdots&1/4^n\\
\vdots
\end{array}
\end{align*}
Now, instead of summing row-by-row, try summing column-by-column:
\begin{align*}
\left(\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+\ldots\right)+\left(\frac{1}{4^2}+\frac{1}{4^3}+\ldots\right)+\left(\frac{1}{4^3}+\ldots\right)+\ldots
\end{align*}
Each summand is of the form $$\sum_{j=k}^{\infty}\frac{1}{4^j}=\frac{1}{4^k}\sum_{j=0}^{\infty}\frac{1}{4^j}=\frac{1}{4^k}\times\frac{1}{1-1/4}=\frac{1}{4^k}\times\frac{4}{3}$$
for $k\in\{1,2,\ldots\}$. Now sum over $k$ to get
$$\sum_{k=1}^{\infty}\frac{1}{4^k}\times \frac{4}{3}=\frac{1/4}{1-1/4}\times\frac{4}{3}=\frac{1}{3}\times\frac{4}{3}=\frac{4}{9}.$$
A: The idea is to take the derivative of the geometric series. For $|x|<1$,
$$\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$
So,
$$\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n.$$
A: Yes, it's just a geometric series of geometric series:
$$
\sum_{n=1}^{\infty}nx^{-n}=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}x^{-n}=\sum_{k=1}^{\infty}\frac{x^{-k}}{1-1/x}=\frac{x^{-1}}{(1-1/x)^2}=\frac{x}{(x-1)^2}.
$$
So your sum is $4 / 9$.
