Prove that the series $\sum_{n=1}^{\infty} \frac{4^n+n^3}{2^n+n!}$ is convergent. can you help me proving that this series is convergent, please?
$$\sum_{n=1}^{\infty} \frac{4^n+n^3}{2^n+n!}$$
I've tried to find a bigger convergent series, but with no luck. 
Thanks.
Lorenzo
 A: HINT
The $n!$ factor in the denominator is stronger than others terms, therefore the series seems prone to converge.
In these cases, as suggested by Lord Shark the Unknown, a good and effective way is choose a simple convergent series (e.g. $\sum \frac1{n^2}$) and proceed by limit comparison test.
In that way the study of the convergence reduces to the calculation of a limit.
A: Ratio test:
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{n\to\infty} \frac{4^{n+1}+(n+1)^3}{4^n+n^3}\cdot \frac{2^n+n!}{2^{n+1}+(n+1)!}=4\cdot 0=0<1.$$
Hence the series converges.
A: I managed to resolve the question. I initially did like @LordSharkTheUnknown and @Gimusi advised, taking $a(n)=\frac{2*4^n}{n!}$ and $b(n) = \frac{1}{4^n}$. For the limit comparison test, since 
$$ \lim_{n \to \infty} \frac{2*4^n*4^n}{n!} = 0$$ and $b(n) $ converges, $\implies a(n)$ converges. Since my initial function is equal or less than $a(n)$, it converges as well.
Than I' ve noticed that also with the ratio test on my bigger series (the one I've called $a(n)$ ) gives $0$ as result, so it converges, but with less effort :) 
Thanks to all of you
