TITLE: Existence of a sequence $\{\epsilon_n\}_{n\ge 1}$ such that $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} $ converges
QUESTION [10 upvotes]: 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 

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

REPLY [9 votes]: Since
$\sum_{n=1}^{\infty}\frac{1}{n\log^2 n}
$
converges,
if we choose
$n^{\varepsilon_n}
=n\log^2 n
$
or
$n^{(\varepsilon_n-1)/2}
=\log n
$
or
$(\varepsilon_n-1)/2
=(\log \log n)/\log n
$
or
$\varepsilon_n
=1+(2\log \log n)/\log n
$,
the resulting series
will converge
and
$\lim_{n \to \infty} \varepsilon_n
= 1
$.
Similarly,
if
$\varepsilon_n
=1+(\log \log n)/\log n
$,
the resulting series
will diverge.