<s>[INST] 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?
 [/INST] 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
$$\sum_{k=1}^{\infty}\frac{1}{4^k}\times \frac{4}{3}=\frac{1/4}{1-1/4}\times\frac{4}{3}=\frac{1}{3}\times\frac{4}{3}=\frac{4}{9}.$$
</s>