Answer
Converges
Work Step by Step
Given
$$ \sum_{m=1}^{\infty}\frac{4}{m !+4^{m}} $$
Compare with $\displaystyle\sum_{m=1}^{\infty}\frac{4}{4^m}$, which is a convergent series ( geometric with $|r|<1$) and for $m\geq 1$
\begin{align*}
m !+4^{m}&\geq 4^{m} \\
\frac{1}{m !+4^{m}} &\leq \frac{1}{4^{m}}\\
\frac{4}{m !+4^{m}}& \leq \frac{4}{4^{m}}
\end{align*}
Then $\displaystyle\sum_{m=1}^{\infty}\frac{4}{m !+4^{m}}$ also converges.
