I'm having difficulties to determine if the series
$$\sum_{n=1}^{\infty}n^{-{(1+1/n)}}$$
converges or not.
I have rewritten the series in the form
$$\sum_{n=1}^{\infty}\frac{1}{n^{(1+1/n)}}=\sum_{n=1}^{\infty}\frac{1}{n\sqrt [n]{n}}$$
but I don't know how I should continue form here.
Hint. Note that $n^{1/n}=e^{\ln n/n}\to 1$.
Use the limit comparison test with $\;a_n:=\cfrac1{n\sqrt[n]n}\;$ , and $\;b_n:=\cfrac1n\;$ :
$$\lim_{n\to\infty}\frac{a_n}{b_n}=1\;\ldots$$
Yet another method. Let $a_n = \frac{1}{n \sqrt[n]{n}}$. Note that $a_n$ is positive and decreasing. Therefore, by Cauchy's Condensation Theorem, one has
$$\sum_{n=1}^{\infty}\frac{1}{n \sqrt[n]{n}} \sim \sum_{n=1}^{\infty}2^n\frac{1}{2^n (2^n)^{\frac{1}{2n}}}=\sum_{n=1}^{\infty}\frac{1}{\sqrt{2}}$$
which clearly diverges.
If you are not familiar with this test look here: Cauchy Condensation Test.
$\frac{1}{2}\sum\limits_{n=1}^N \frac{1}{n}\leq \sum\limits_{n=1}^N \frac{1}{n\sqrt[n]{n}}\leq \sum\limits_{n=1}^N \frac{1}{n}$ because of $1\leq n<2^n$
and $\sum\limits_{n=1}^N \frac{1}{n}$ is divergent for $N\to\infty$
