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!
 A: With an equivalent:
$$
\frac{1}{n^{(n+1)/n}} = \exp\left(-\left(1+\frac{1}{n}\right)\ln n\right) = e^{-\ln n+o(1)} \sim \frac{1}{n}.
$$

With an inequality (using $\ln(1+x) \leq x$):
$$
\dfrac{n}{n^{(n+1)/n}} = \exp\left(-\frac{\ln n}{n}\right) \geq \exp\left(-\frac{n-1}{n}\right) = \exp\left(\frac{1}{n}-1\right) \geq e^{-1}.
$$
Hence, $$\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^{(n+1)/n}} \geq \dfrac{1}{e} \sum_{n=1}^\infty\dfrac{1}{n} = +\infty.$$
A: Method 1:  Limit comparison with $\frac{1}{n}$
Method 2: $$\frac{1}{n^{1+1/n}}\leq\frac{1}{(n+1)^{1+\frac{1}{n+1}}}$$ Now use Cauchy condensation test, to get that syor sum converges iff the following sum converges:
$$\sum_{n\in Z^+}2^n\frac{1}{(2^n)^{1+2^{-n}}}=\sum_{n\in Z^+}\frac{1}{2^{n2^{-n}}}$$
Since $n<2^n$, therefore $n2^{-n}<1$. Thus, $\frac{1}{2}<\frac{1}{2^{n2^{-n}}}$
A: 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. 
