Find $\sum\limits_{n=1}^{\infty}\frac{n^4}{4^n}$ So, yes, I could not do anything except observing that in the denominators, there is a geometric progression and in the numerator, $1^4+2^4+3^4+\cdots$. 
Edit: I don't want the proof of it for divergence or convergence only the sum.

\begin{align}
n^4 = {} & \hphantom{{}+{}} An(n-1)(n-2)(n-3) \\ 
& {} + B n(n-1)(n-2) \\
& {} + C n(n-1) \\
& {} + D n
\end{align}
Find $A,B,C,D$. Then
\begin{align}
n^4 x^n & = An(n-1)(n-2)(n-3) x^n + \cdots \\[10pt]
& = A\frac{d^4}{dx^4} x^n + B \frac{d^3}{dx^3} x^n + \cdots
\end{align}
Then add over all values of $n$, and then find the derivatives.
Finally, apply this to the case where $x=\dfrac 1 4$.
