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?
 A: TO MAKE A SENCE $$a=\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...$$multiply by $\frac 12$
$$\frac12a=\frac{0}{4}+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\frac{5}{128}+...$$ now notice to $a-\frac 12 a$
$$a-\frac 12 a=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}-\frac 1{8}+\frac{3}{16}-\frac{2}{16}+\frac{4}{32}-\frac{3}{32}+\frac{5}{64}-\frac{4}{64}+...=\\
\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...=geometric$$
$$\frac12a=\frac{0}{4}+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\frac{5}{128}+...=0+\sum^{\infty}_{k=2}\frac{k-1}{2^{k+1}}\\**k-1=u**\\=0+\sum^{\infty}_{u=1}\frac{u}{2^{u+2}}$$and 
$$a=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...=0+\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{u+1}{2^{u+2}}}_{\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...}$$  now apply $a-\frac12 a$
$$a-\frac12 a=\\\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{u+1}{2^{u+2}}}_{\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...}-\sum^{\infty}_{u=1}\frac{u}{2^{u+2}}=\\
\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{\not u+1-\not u}{2^{u+2}}}_{geometric}$$
A: 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}
$
A: 
if you have $x^3$ in the denominator you end up with $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}$

That's only true when there are other factors in the denominator, such as $(x-1)$. When you have no other factors, and have just $x^3$ in the denominator, there is no simplification to be made, you already have the simplest form.
Similarly in this case, there isn't any way to simplify this sum using partial fractions, nor is there a way to express the summand in partial fractions.

With regards to the actual question, consider what happens when you integrate w.r.t. $x$ $$\sum_{k=1}^\infty\frac{k}{x^{k+1}}$$
