Convergence of the series $\sqrt[n]n-1$

Hint: $$\sqrt[n]n = e^{\frac{\log n}{n}} > 1+\frac{\log n}{n}$$

$$a_n=n^{1/n}-1= e^{\log{n}/n}-1 \sim \frac{\log{n}}{n}$$

The sum diverges by comparison with the sum of $b_n=1/n$.

From $\left(1+\frac1n\right)^n\to e$, we conclude $$ \left(1+\frac1n\right)^n<3<n=(1+a_n)^n$$ for almost all $n$. Hence $a_n>\frac1n$ for almost all $n$ and $\sum a_n$ diverges.

Remark: We don't even need the introduction of $e$. The observation $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}n^{-k}\le \sum_{k=0}^n\frac1{k!}<1+\sum_{k=1}^ \infty 2^{1-k} =3$$ suffices for an "elementary" approach.