Sequence sum question: $\sum_{n=0}^{\infty}nk^n$ I am very confused about how to compute 
$$\sum_{n=0}^{\infty}nk^n.$$
Can anybody help me?

By the ratio test your series converges for $|k|\lt 1$. Let $|k|\lt 1$, consider $$S_m=\sum_{n=0}^m nk^n.$$
Then
$$\begin{align*}
S_m &= 0 + k + \sum_{n=2}^m nk^n\\
&= k +\sum_{n=1}^{m-1}(n+1)k^{n+1}\\
&= k + k\sum_{n=1}^{m-1} nk^n + \sum_{n=1}^{m-1} k^{n+1}\\
&= k + k\sum_{n=1}^{m-1} nk^n + k^2\sum_{n=1}^{m-1} k^{n-1}\\
&= k + kS_{m-1} + k^2\sum_{n=0}^{m-2} k^n\\
&= k + kS_{m-1} + k^2\cdot \frac{1-k^{m-1}}{1-k}.
\end{align*}$$
We know that $S_m\to l$ for some $l\in\mathbb{R}$. Taking limits in the last equality we get
$$\begin{align*}
\lim_{m\to\infty}S_m &= k+k\lim_{m\to\infty} S_{m-1}+\frac{k^2}{1-k}-\lim_{m\to\infty}\frac{k^{m+1}}{1-k}\\
l &= k+kl+\frac{k^2}{1-k}-0\\
(1-k)l &= k+\frac{k^2}{1-k}\\
l&= \frac{k}{1-k}+\left( \frac{k}{1-k} \right)^2\\
l&=\frac{k}{(1-k)^2}.
\end{align*}$$
