If $A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}$, find $\lim_{n \to \infty} A_n$ 
Set $$A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}.$$ Find $\displaystyle \lim_{n \to \infty} A_n$.

I tried solving this using a reduction method. That is, reducing the above series to an Arithmetico-Geometric series. We can see this by noting that $$\sum_{k = 1}^{\infty} \dfrac{k^m}{2^k} = \dfrac{2}{3} \left(1+\dfrac{2^m}{2} +\sum_{k=1}^{\infty} \dfrac{(k+1)^m-(k-1)^m}{2^k}\right)$$ which can be derived using infinite series. Using this method, we can eventually reduce the series to something of the form $\displaystyle C\sum_{k=1}^{\infty}  \dfrac{k}{2^k}$ for some constant $C$ which is easily evaluated. Although this method would require a minimum of six iterations which would be quite lengthy. I am wondering if there is a more efficient way of solving this.  

Note that $$\sum_{k=1}^\infty x^k = \frac{1}{1-x} - 1$$ converges uniformly on every compact subset of $(-1,1)$ so that 
$$\sum_{k=1}^\infty k x^{k-1} = \frac{1}{(1-x)^2}$$ by differentiation and
$$\sum_{k=1}^\infty kx^k = \frac{x}{(1-x)^2}$$ after multiplication by $x$.
Repeat five times and plug in $x=\frac 12$.
