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.

$$\sum_{n\geq 1}\frac{e^{nx}}{4^n} = \frac{e^x}{4-e^x} $$
since it is the sum of a geometric progression. Differentiating four times with respect to $x$,
$$\sum_{n\geq 1}\frac{n^4}{4^n}e^{nx} = \frac{256 e^x+704 e^{2 x}+176 e^{3 x}+4 e^{4 x}}{(4-e^x)^5}$$
and by evaluating at $x=0$,

$$\sum_{n\geq 1}\frac{n^4}{4^n}= \color{red}{\frac{380}{81}}.$$

