Evaluate $\sum_{n=1}^\infty \frac{n}{2^n}$ This is a homework question; I'm supposed to use power series to find the following sum: $$\sum_{n=1}^\infty \frac{n}{2^n}$$ I took the geometric series $$\frac{1}{1-x}=\sum_{n=0}^\infty {x^n}$$ and differentiated and multiplied both sides by x to get $$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty {nx^n}$$ I'm stuck because I'm not sure how to make the $$\frac{1}{2^n}$$ term appear.
 A: For $x=\frac {1} {2}$, we have $\sum _{i} nx^n=\sum _{i} n(\frac {1} {2})^n=\sum _{i} \frac {n} {2^n}$
A: Without power series:
$$2S-S:=\frac11+\frac22+\frac34+\frac48\cdots-\frac12-\frac24-\frac38-\cdots
=\frac11+\frac12+\frac14+\frac1{8}+\cdots=2.$$
A: You are basically all of the way there.  In your last step, you have
$$ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n, \tag{$\ast$}$$
where the series on the right converges for any $|x| < 1$ to the rational expression on the left.  Since we get to pick $x$ to be anything we like (as long as $|x| < 1$), we might as well choose $x = \frac{1}{2}$.  If we do this, the series on the right-hand side of ($\ast$) becomes
$$ \sum_{n=1}^{\infty} n \left( \frac{1}{2} \right)^n
= \sum_{n=1}^{\infty} n \cdot \frac{1}{2^n}
= \sum_{n=1}^{\infty} \frac{n}{2^n}, $$
which is the original series that you were given to evaluate.  But with $x = \frac{1}{2}$, the left-hand side of ($\ast$) is
$$ \frac{1}{\left( 1 - \frac{1}{2} \right)^2} = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4.$$
Therefore, equating the left- and right-hand sides of ($\ast$) when $x = \frac{1}{2}$, we obtain
$$ \sum_{n=1}^{\infty} \frac{n}{2^n} = 4.$$
