How Can I prove $\sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$ Question:Prove that $ \sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$
While doing questions on series and products,i got stuck in this question.I was not able to figure out any way how to prove this one.Convergence tests shows that this series converges but i don't know how to find its value.
 A: Here's a hint:
You should be able to evaluate the series
$$\sum_{n=1}^\infty \frac{1}{5^n}$$
with a simple geometric series argument. More generally, this allows you to say something like
$$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$
for $-1<x<1$. The magic happens when you take the derivative of this to get
$$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2},$$
which gives that
$$\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}.$$
However, we want a higher power of $x$. What happens if we take the derivative again? Can you generalize this enough to solve your problem?
A: Equivalently, we want to evaluate $\left.\sum_nn^4x^n\right|_{x=1/5}$ with $\sum_n:=\sum_{n=\color{blue}{0}}^\infty$. By the binomial theorem, if $|x|<1$ then$$\begin{align}(1-x)^{-1}&=\sum_nx^n,\,\\(1-x)^{-2}&=\sum_n(n+1)x^n,\,\\(1-x)^{-3}&=\sum_n\tfrac12(n+1)(n+2)x^n,\,\\(1-x)^{-4}&=\sum_n\tfrac16(n+1)(n+2)(n+3)x^n,\,\\(1-x)^{-5}&=\sum_n\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)x^n.\end{align}$$We'll find a linear combination by first getting the $n^4$ coefficient right, then $n^3$ etc. Since$$\begin{align}n^4&=24\cdot\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)\\&-60\cdot\tfrac16(n+1)(n+2)(n+3)\\&+50\cdot\tfrac12(n+1)(n+2)\\&-15\cdot(n+1)\\&+1,\end{align}$$we have$$\begin{align}\sum_nn^4x^n&=24(1-x)^{-5}-60(1-x)^{-4}\\&+50(1-x)^{-3}-15(1-x)^{-2}+(1-x)^{-1}.\end{align}$$Substituting $x=\tfrac15$ gives the desired result.
