Cauchy Criterion divergence I have problem checking of this series converge or diverge using Cauchy Criterion:

$$ \sum \frac {\sin(n^n+e^{n!})}{4^n} $$

I managed to show that the sum is smaller then $ \frac{p}{4^n}$ since $ -1\le \sin(x) \le 1 $ but I think I have to get rid of $p$ somehow.

Any advice?
Cauchy Criterion for convergence:

For every $\enspace\epsilon&gt;0\enspace$ exists $\enspace N&gt;0\enspace$ with $\enspace\displaystyle |\frac{\sin(m^m+e^{m!})}{4^m}-\frac{\sin(n^n+e^{n!})}{4^n}|&lt;\epsilon$ 

for all $\enspace m,n\ge N$ .

Proof:

For $\enspace N\leq \min(m,n)\enspace$ and $\enspace\displaystyle \epsilon:=\frac{1}{4^{N-1}}\enspace$ we get  

$\displaystyle |\frac{\sin(m^m+e^{m!})}{4^m}-\frac{\sin(n^n+e^{n!})}{4^n}|\leq \frac{1}{4^m}+\frac{1}{4^n}\leq \frac{2}{4^{\min(n,m)}}&lt;\frac{1}{4^{N-1}}=\epsilon\enspace$ .
We have $|\frac {sin(n^n+e^{n!})}{4^n}| \le \frac {1}{4^n}$ for all $n$ and $\sum\frac {1}{4^n}$ is convergent ......