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 

$$\sum_{n= 1}^{\infty}  \frac{1}{n^1}=\infty$$
 A: 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 $n\to \infty$. 
A: 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.
A: Thoughts, but definitely not an answer: 
A standard proof that $\sum \frac{1}{n}$ diverges is to group things in groups that are size a power of $2$. (1) + (1/2 + 1/3) + (1/4 + ... + 1/8) + ... and then argue that each element in a list is at least as big as the last one, and therefore each list has a sum that's at least $1/2$. 
A kind of "dual" of that is to say that each element is no \emph{larger} than the first, and therefore each group is no more than $2^k$ times the first element (where the group has size $2^k$). If only those numbers ($2^k$ times the first element) were well-behaved, this would get you an UPPER sum for the series and you could show it was bounded. It's not, of course, but there's half an idea there. 
Suppose your numbers $\epsilon_n$ approached $1$ from above slowly enough that each group (in some grouping like the one above) had a sum that was manageable. Then you could actually prove boundedness. One thing you realize is that the decrease in your exponents has to be more or less exponentially slow. 
