<s>[INST] 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?
 [/INST] 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 $a-\frac12 a$
$$a-\frac12 a=\\\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{u+1}{2^{u+2}}}_{\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...}-\sum^{\infty}_{u=1}\frac{u}{2^{u+2}}=\\
\frac{1}{4}+\underbrace{\sum^{\infty}_{u=1}\frac{\not u+1-\not u}{2^{u+2}}}_{geometric}$$
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