How to simplify this summation: $\sum\limits_{k=0}^\infty \frac1{4!} \cdot \frac{k^7}{2^k}$

Question

I was wondering how to solve this infinite sum.
$$\sum_{k=0}^\infty {1\over 4!} \cdot {k^7\over2^k}$$ I know roughly that for $$\sum_{k=0}^\infty {k \over 2^k}$$ the sum takes advantage of the derivative of $(1-x)^{-1}$ to get the result, but I'm not full clear on that as well as how to extend it to the $k^7$ properly. Or is there a better way to go about this problem?

Answer 1

Knowing that $~\displaystyle\sum_{k=0}^\infty x^k=\dfrac1{1-x}$ , all that's left to do is to apply the following two operations seven times with regard to x, in exactly this order: differentiation, and multiplication. Then let $x=\dfrac12$.

Answer 2

Note that $$ \sum_{k=0}^{+\infty}x^k = \dfrac{1}{1 - x} \quad x \in (-1,1) $$ Let differential operator $D = \dfrac{d}{dx}$. Hence, $$ D\sum_{k=0}^{+\infty}x^k = \sum_{k=1}^{+\infty}kx^{k-1}= D\biggl(\dfrac{1}{1 - x}\biggr) \quad \Rightarrow \quad \sum_{k=1}^{+\infty}kx^k = xD\biggl(\dfrac{1}{1 - x}\biggr) $$ By finite induction and taking $p \in \mathbb{N}$, $$ \sum_{k=p}^{+\infty}k^px^k = (xD)^p\biggl(\dfrac{1}{1 - x}\biggr) $$ For $p =7$ and $x = 1/2$, we have $$ \sum_{k=7}^{+\infty}\dfrac{k^7}{2^k} = (xD)^7\biggl(\dfrac{1}{1 - x}\biggr)\biggl|_{x=1/2} $$