Test for convergence the series $\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$ Test for convergence the series
$$\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$$
I'd like to make up a collection with solutions for this series, and any new
solution will be rewarded with upvotes.   Here is what I have at the moment
Method 1 
We know that for all positive integers $n$, $n<2^n$, and this yields
$$n^{(1/n)}<2$$
$$n^{(1+1/n)}<2n$$
Then, it turns out that 
$$\frac{1}{2} \sum_{n=1}^{\infty}\frac{1}{n} \rightarrow \infty \le\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$$
Hence
$$\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}\rightarrow \infty$$
EDIT: 
Method 2
If we consider the maximum of $f(x)=x^{(1/x)}$ reached for $x=e$
and denote it by $c$, then
$$\sum_{n=1}^{\infty}\frac{1}{c \cdot n} \rightarrow \infty \le\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$$
Thanks!

Let $a_n = 1/n^{(n+1)/n}$. 
Then 
$$\begin{eqnarray*}
\frac{a_n}{a_{n+1}} &\sim& 1+\frac{1}{n} - \frac{\log n}{n^2}
\qquad (n\to\infty).
\end{eqnarray*}$$
The series diverges by Bertrand's test. 
