Does the following fractional series converge? Does $\sum_{n=1}^{\infty}(\frac{5^{2n+1}}{n^n})$ converge?. So far I believe it does not since numerator seems to me to grow faster but I was not able to write the proof, so any hint is very appreciated.
 A: Yes, it does. A standard technique to show convergence is to bound a series term-wise by a bigger series. In this case, we bound the series from above with a geometric series (after discarding the first few terms). Let $$ S = \sum_{n=1}^\infty \left(\frac{5^{2n+1}}{n^n}\right). $$
We then have
\begin{align*}
S &= 5\cdot\sum_{n=1}^\infty \left(\frac{25}{n}\right)^n \\
&\le 5\left(\sum_{n=1}^{25} \left(\frac{25}{n}\right)^n + \sum_{n=26}^{\infty} \left(\frac{25}{26}\right)^n\right).
\end{align*}
The first term is clearly finite; the second term is a geometric series that converges. Hence $S$ converges as well.
A: Considering $$u_n=\frac{5^{2n+1}}{n^n}$$ we have $$r_n=\frac{u_{n+1}}{u_n}=25 \frac {n^n} {(n+1)^{n+1}}$$ So, $$\log(r_n)=\log(25)+n\log(n)-(n+1)\log(n+1)$$ what you can write as $$\log(r_n)=\log(25)+n\log(n)-n\left(1+\frac 1n\right)\left(\log(n)+\log\left(1+\frac 1n\right)\right)$$ Now, by Taylor for large $n$ , $\log\left(1+\frac 1n\right)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$ $$\log(r_n)=\log(25)-1-\log(n)+O\left(\frac{1}{n^2}\right)\implies r_n=\frac{25}{e n}+O\left(\frac{1}{n^2}\right)$$ from which you can conclude.
