Simplify the sum $ \sum_{k=1}^{\infty} (\frac{1}{2})^kk $

Question

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.

Answer 1

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 $f(1/2)$.

Answer 2

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

$$S=1+1/2+1/4+1/8+\ldots = 2.$$

Answer 3

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 $$ \sum_{k=1}^{\infty} \left(\tfrac{1}{2}\right)^kk = \sum_{k=0}^{\infty} \Big(\tbinom {k+1}1-\tbinom k0\Big) \left(\tfrac{1}{2}\right)^k = (1-\tfrac12)^{-2}-(1-\tfrac12)^{-1} = 4-2=2. $$