<s>[INST] How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n} $ I have put his on wolfram and obtained answer as follows:
$\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$
And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$
However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?
 [/INST] In this answer we go for the 'one trick pony' approach - all we know is that for $k \ge 0$,
$\tag 1 \sum_{n=k}^{\infty}\frac{1}{2^n} = 2^{1-k}$
We decompose the summands of $\sum_{n=1}^{\infty}\frac{n}{2^n}$ in a natural/straightforward manner, arranging these numbers into a table:
$$\begin{pmatrix}
\frac{1}{2} & \frac{1}{4}  & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & \frac{1}{4}  & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & 0  & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & 0  & 0 & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & 0 & 0 & 0 & \frac{1}{32} & \dots \\
0 & 0 & 0 & 0 & 0 & \dots \\
. \\
. \\
. \\
\end{pmatrix}$$
Using $\text{(1)}$ we add up the entries in each row,
$$\begin{pmatrix}
1  \\
\frac{1}{2} \\
\frac{1}{4} \\
\frac{1}{8} \\
\frac{1}{16} \\
. \\
. \\
. \\
\end{pmatrix}$$
And now we add up the entries of our column vector, giving
$\quad \text{ANS: } 2$
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