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?

Let $S$ be our sum. Then $S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$. Thus
$$\begin{align}2S&=1+&\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\cdots\\
S&=&\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots\end{align}
$$
Subtract. We get
$$S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=2.$$
