Existence of a sequence $\{\epsilon_n\}_{n\ge 1}$ such that $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} $ converges I am trying to find a sequence $\{\varepsilon_n\}_{n\ge 1}$  such that
$$\lim_{n\to \infty}\varepsilon_n=1~~~~~~~~~~~~~\text{and}~~~~~~~~~\sum_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} <\infty.$$
In case such sequence $\{\varepsilon_n\}_{n\ge 1}$ exists I would like to have an explicit example.

Remark: This is an interesting problem since we know that for the case where  $\varepsilon_n = 1$ we have 

$$\sum_{n= 1}^{\infty}  \frac{1}{n^1}=\infty$$

Yes, you can find such $\epsilon_n$. First, the series below is convergent 
$$
\sum_{n\geq 2} \frac{1}{n (\ln n)^{1+s}}
$$
for any $s>0$.  We choose $s=1$ and $\epsilon_n$ such that $2n(\ln n)^2>n^{\epsilon_n}>n(\ln n)^2$ for any $n\geq 2$. We can find that 
$\epsilon_n\to 1$ as $n\to \infty$. 
