using the ratio test
$L=\lim \limits_{n\to \infty}~\dfrac{\dfrac{e^{n+1}+(n+1)}{(n+1)^5}}{ \dfrac{e^n+n}{n^5}} = $
$L=\lim \limits_{n\to \infty}~\dfrac{n^5(e^{n+1}+n+1)}{(n+1)^5(e^n+n)}$
$L = e > 1$ and thus the series is divergent
should be pretty clear that $\dfrac{n^5}{(n+1)^5}$ goes to 1
should also be pretty clear that the $e^n$ term leaves the $n$ and constant terms in the dust so this becomes $\dfrac{e^{n+1}}{e^n} = e$
you can show all that formally but that's the common sense explanation.
I offer another approach that uses the comparison & root test.
$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{{e^n} + n}}{{{n^5}}}} \ge \sum\limits_{n = 1}^\infty {\frac{{{e^n}}}{{{n^5}}}} $
But
$\displaystyle \sqrt[n]{{\frac{{{e^n}}}{{{n^5}}}}} = \frac{e}{{{{\left( {\sqrt[n]{n}} \right)}^5}}} \to e > 1$
This shows divergence.
Thank you ! this approach seems a lot more simpler, however I fail to understand how to make the bottom part (nsquarerootn)^5 go to 1 for it to be e/1. any tricks to simplify I could use? I pretty much tried to make it (n^(1/n))^5 but still dont see how it goes to 1, since my teacher said we cant just assume that infinity^(0) gives 1.