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.

Let 
$$f(x) = \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$
Then the sum is
$$\left [\left (x \frac{d}{dx} \right )^4 f(x)\right ]_{x=1/4}$$
