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?
 A: 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
$$S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=2.$$
A: Hint:
$$ \sum_{k=1}^{\infty} x^k = \frac{x}{1-x} \implies \sum_{k=1}^{\infty} kx^k=\frac{d}{dx} \frac{x}{1-x} .  $$
A: Another way to do the problem is to note that
$$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 = \sum_{k=1}^\infty \frac{k-1}{2^k}$$
so that the desired series is really just
$$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 + \sum_{k=1}^\infty \frac{1}{2^k}$$
A: If $\displaystyle S=\sum_{n=1}^\infty\frac{n}{2^n}$, then $$S-\frac12=\sum_{n=1}^\infty\frac{n+1}{2^{n+1}}=\frac12\sum_{n=1}^\infty\frac{n+1}{2^n}=\frac12\left(S+\sum_{n=1}^\infty\frac1{2^n}\right)=\frac12(S+1),$$ so $2S-1=S+1$, or $S=2$. 
All that remains is to justify that the series converges. But $n<1.1^n$ for $n$ large enough, so a tail of $S$ is bounded above by a tail of the convergent geometric series $\sum_n\left(\frac{1.1}2\right)^n$.
A: 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
$\quad \text{ANS: } 2$
