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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.

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  • $\begingroup$ $m^m=m\cdot m\cdot m \dotsb m > m^2\cdot (1\cdot 2\cdot 3\dotsb m) / 2=\frac12m^2\cdot m!$. $\endgroup$ – W-t-P Feb 25 at 19:52
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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.

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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.

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