## Calculus (3rd Edition)

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.