Answer
The series diverges by limit comparison.
Work Step by Step
$\Sigma_{n=1}^{\infty}\frac{n^{k-1}}{n^k+1}, k\gt2$
Let's compare the original series with the series $\Sigma_{n=1}^{\infty}\frac{n^{k-1}}{n^k}=\Sigma_{n=1}^{\infty}\frac{1}{n}$, which diverges by p-series, where $p=1$, $0\lt1\leq1$
Now let's use limit comparison:
If $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=L$, where $L$ is both finite and positive, then either both series diverge or both series converge, depending on the convergence or divergence of $b_n$:
$\lim\limits_{n \to \infty}\frac{n^{k-1}}{n^k+1}\times\frac{n^k}{n^{k-1}}$
$\lim\limits_{n \to \infty}\frac{n^k}{n^k+1}=\frac{\infty}{\infty}$
$\frac{\infty}{\infty}$ is an indeterminate form, so let's use L'Hopital's rule:
$\lim\limits_{n \to \infty}\frac{kn^{k-1}}{kn^{k-1}}=1$
Because $1$ is a finite and positive number, by limit comparison we can conclude that the original series also diverges.
