Proving that $\sum_{m=1}^\infty \frac{m!}{2^m+m^m}$ converges

Question

I'm trying to prove that $\sum_{m=1}^\infty \frac{m!}{2^m+m^m}$ converges. I've only been capable of doing this via the (limiting) ratio test where I've proved that $lim_{m \rightarrow \infty} \frac{a_{m+1}}{a_m} = \frac{1}{e} < 1$. However, my method is rather cumbersome and I'm wondering if there's a simpler method to prove this using the comparison or ratio test.

Answer 1

The $m$th term is bounded above by $\dfrac{m!}{m^m}.$ But note that if $m\ge 2,$ then

$$\frac{m!}{m^m} = \frac{m}{m}\cdot \frac{m-1}{m}\cdots \frac{2}{m}\cdot\frac{1}{m} \le \frac{2}{m^2}.$$

Since $\sum 1/m^2$ converges, so does the given series.

Answer 2

You can notice that $$\frac{m!}{2^m+m^m}\leq \frac{m!}{m^m}$$ so that if $\sum\frac{m!}{m^m}$ converges, then surely $\sum \frac{m!}{2^m+m^m}$ converges. But $$\lim_{m\to\infty} \frac{\frac{(m+1)!}{(m+1)^{m+1}}}{\frac{m!}{m^m}} = ... = \frac{1}{e} < 1 $$so the series converges. I suggest this method because calculations might be easier for you.