Infinite sum of $nk^n$ How do you evaluate a sum of the form $\sum\limits_{n=0}^\infty nk^n$, for $|k|<1$?
If we take $k=\frac{1}{2}$, the sum is $\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \dots$, which isn't like any series I know. Wolfram Alpha evaluates this to $\frac{k}{(k-1)^2}$, but I don't know how it finds that.
 A: The power series
$$ \frac{1}{1 - z} = \sum_{n \geq 0} z^n$$
converges absolutely for all $|z| < 1$. Some results about power series also say that it will be well-behaved under differentiation in that region, so differentiate both sides.
$$ \frac{1}{(1-z)^2} = \sum_{n \geq 0} n z^{n-1}$$
which is almost, but not quite, the series you want. Multiplying both sides by $z$ fixes this:
$$ \frac{z}{(1 - z)^2} = \sum_{n \geq 0} n z^n$$
which is the closed form that Wolfram Alpha gave.
A: Consider $S(n)=\sum_{n=0}^{\inf}k^n$. If you take the derivative of $S(n)$ (we are allowed to do this because this is a power series with $|k|<1$), you will notice that you get another infinite sum with the slight modification: $$K(n)=\sum_{n=0}^{\inf}(n+1)k^n$$. Notice that the sum of your interest, $$L(n)=k.K(n).$$ Since $S(n)$ is an infinite geometric series that converges, you can calculate this explicitly and that will lead you to your answer.
Or you can choose to use induction since you know the answer. 
A: If you don't want to use derivative of power series:
\begin{align}
\frac{z-z^{n+1}}{1 - z}& =& &z+&z^2+&z^3+...+&z^n \\
\frac{z^2-z^{n+1}}{1 - z}& =& &&z^2+&z^3+...+&z^n \\
\frac{z^3-z^{n+1}}{1 - z}& =& &&&z^3+...+&z^n \\
\frac{z^n-z^{n+1}}{1 - z}& =& &&&&z^n \\
\end{align}
Adding them together you get
$$\sum_{k=1}^n kz^k = \frac{z+z^2+..+z^n-nz^{n+1}}{1-z}= \frac{z\frac{1-z^{n}}{1-z}-nz^{n+1}}{1-z}=\frac{z-(n+1)z^{n+1}+nz^{n+2}}{(1-z)^2}$$
Now, use the fact tha for $|z|<1$ we have
$$\lim_n (n+1)z^{n+1}=\lim_n nz^{n+2}=0$$
