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.

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
$$\lim_n (n+1)z^{n+1}=\lim_n nz^{n+2}=0$$
