Simplify the sum $ \sum_{k=1}^{\infty} (\frac{1}{2})^kk $ I need some help simplifying this sum:
$$ \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^kk $$
I have a feeling it's some basic series thing that I'm forgetting, but I need help nonetheless. 
 A: Put $$f(y)=\sum_{x=1}^{\infty} xy^x=y\sum_{x=1}^{\infty} xy^{x-1}=y\left(\sum_{x=1}^{\infty} y^x\right)'=y\left(\frac{1}{1-y}\right)'$$
Compute the right-hand side above and then put $f(1/2)$.
A: Let $S$ be the sum.  If we multiply $S$ by $1/2$ and subtract from $S$, we have
$$ S/2 = (1\cdot (1/2) + 2\cdot (1/2)^2 +\ldots)-(0\cdot (1/2) + 1\cdot(1/2)^2  + 2\cdot (1/2)^3+\ldots)$$
Regrouping terms based on the power of $(1/2)$, we have
$$ S/2 = (1-0)(1/2)+(2-1)(1/2)^2+(3-2)(1/2)^3+\ldots$$
and so 
$$S=1+1/2+1/4+1/8+\ldots = 2.$$
A: I'll try to make this an instance of a more general method, that will work for expressions where the coefficient $k$ is replaced by any polynomial expression of$~k$.
You can recognise this as an instance of the binomial formula with negative exponent. In general
$$
  (1+x)^{-n} = \sum_{k=0}^\infty\binom{-n}kx^k
  \qquad\text{where }
  \binom{-n}k=(-1)^k\binom{k+n-1}k = (-1)^k\binom{k+n-1}{n-1}
$$
for $|x|<1$, which because of the factor $(-1)^k$ is more pleasantly represented as 
$$
  (1-x)^{-n}= \sum_{k=0}^\infty\binom{k+n-1}{n-1}x^k.
$$
Here the coefficient $\binom{k+n-1}{n-1}=\frac{(k+n-1)(k+n-2)\ldots k}{(n-1)!}$ grows as a polynomial function of $k$, of degree $n-1$. It the given expression $k$ is a polynomial expression of degree $1$, so you can try $n=2$, but this gives $\binom{k+n-1}{n-1}=\binom{k+1}1=k+1$ instead. So you need to subtract the constant function $1$ of $k$, which you can write as $1=\binom k0$, the formula you get for $n=1$. The rest is now simple rewriting
$$
  \sum_{k=1}^{\infty} \left(\tfrac{1}{2}\right)^kk
= \sum_{k=0}^{\infty} \Big(\tbinom {k+1}1-\tbinom k0\Big)
             \left(\tfrac{1}{2}\right)^k
= (1-\tfrac12)^{-2}-(1-\tfrac12)^{-1}
= 4-2=2.
$$
