how to prove $\sum {\frac{1}{n^{1+1/n}}}$ is divergent

We know that $\sum \frac{1}{n^p}$ is convergent for $p>1$. However the series $\sum {\frac{1}{n^{1+1/n}}}$ is apparently divergent since $1+1/n$ tends to 1 as $n$ tends to infinity. But how to prove this? The root test fails expectedly and I haven't been able to find a smaller divergent series for comparison test. Can someone help? Thanks!

Solution 1:

The limit comparison test:

$$\frac{\frac{1}{n^{1+1/n}}}{\frac{1}{n}}=\frac{1}{\sqrt[n]n}\xrightarrow[n\to\infty]{}1$$

Thus, the series $\,\displaystyle{\sum_{n=1}^\infty\frac{1}{n\sqrt[n] n}\;\;,\;\;\sum_{n=1}^\infty\frac{1}{n}}\,$ converge/diverge together...

Solution 2:

Note that $n^{1/n} \lt 2$ for all $n$, for it is easy to show that $n \lt 2^n$. One can do this by induction, or by using the Binomial Theorem on $(1+1)^n$, or in several other ways.

It follows that $\dfrac{1}{n^{1+1/n}}\gt \dfrac{1}{2n}$.