convergence of $\sum_{n≥ 1}\left(n^{-1-\frac{1}{n}}\right)$ I want to determine the convergence of the series and to do so, consider that
$$\sum_{n≥ 1}\left(n^{-1-\frac{1}{n}}\right)=\sum_{n≥ 1}\left(n^{\frac{-n-1}{n}}\right)$$
I consider at first glance that the series converges, and to determine this I was thinking of using the limit comparison test; however I would have to find a sequence $b_n$ such that $\frac{a_n}{b_n}=L≥0$ Clearly where $b_n$ converges.
Any help? or any other way to approach this? thanks.

You have
$$
n^{-1-\frac{1}{n}} = \frac{1}{n\cdot n^{\frac{1}{n}}} = \frac{1}{n \cdot e^{\frac{\ln n}{n}}} 
$$
Now, what is the limit of $e^{\frac{\ln n}{n}}$, and can you use this to conclude?
