How to do partial fractions with a denominator to the power of a variable? How can I take the following sum and simplify it with partial fractions? $$\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}$$ 
I know the denominator can be rewritten as $(2^k)(2)$, but how do I deal with the $2^k$ when doing partial fractions?  
Usually, when there is an exponent in the denominator, you put terms all the way through - if you have $x^3$ in the denominator you end up with $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}$. So, what do I do when I have $2^k$ - and I either do not know what $k$ or $k$ goes to infinity?

Here's a recurrence for
these power sums.
$\begin{array}\\
s_m(x)
&=\sum_{n=0}^{\infty} n^mx^n\\
&=\sum_{n=1}^{\infty} n^mx^n\\
&=x\sum_{n=1}^{\infty} n^mx^{n-1}\\
&=x\sum_{n=0}^{\infty} (n+1)^mx^{n}\\
&=x\sum_{n=0}^{\infty} x^{n}\sum_{k=0}^m \binom{m}{k}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}\sum_{n=0}^{\infty} x^{n}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}\sum_{n=1}^{\infty} x^{n}n^k\\
&=x\sum_{k=0}^m \binom{m}{k}s_k(x)\\
&=x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)+xs_m(x)\\
\end{array}
$
so
$\begin{array}\\
(1-x)s_m(x)
&=x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)\\
\end{array}
$
or
$\begin{array}\\
s_m(x)
&=\dfrac{x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)}{1-x}\\
s_0(x)
&=\dfrac{1}{1-x}\\
s_1(x)
&=\dfrac{x\sum_{k=0}^{m-1} \binom{m}{k}s_k(x)}{1-x}\\
&=\dfrac{x\dfrac{1}{1-x}}{1-x}\\
&=\dfrac{x}{(1-x)^2}\\
s_2(x)
&=\dfrac{x\sum_{k=0}^{1} \binom{2}{k}s_k(x)}{1-x}\\
&=\dfrac{x(s_0(x)+2s_1(x))}{1-x}\\
&=\dfrac{x(\dfrac{1}{1-x}+2\dfrac{x}{(1-x)^2})}{1-x}\\
&=\dfrac{x((1-x)+2x)}{(1-x)^3}\\
&=\dfrac{x(1+x)}{(1-x)^3}\\
s_3(x)
&=\dfrac{x\sum_{k=0}^{2} \binom{3}{k}s_k(x)}{1-x}\\
&=\dfrac{x(s_0(x)+3s_1(x)+3s_2(x))}{1-x}\\
&=\dfrac{x(\dfrac{1}{1-x}+3\dfrac{x}{(1-x)^2}+3\dfrac{x(1+x)}{(1-x)^3})}{1-x}\\
&=\dfrac{x((1-x)^2+3x(1-x)+3x(1+x))}{(1-x)^4}\\
&=\dfrac{x(1-2x+x^2+3x-3x^2+3x+3x^2)}{(1-x)^4}\\
&=\dfrac{x(1+4x+x^2)}{(1-x)^4}\\
\end{array}
$
