Existence of a sequence $\{\epsilon_n\}_{n\ge 1}$ such that $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\varepsilon_n}} $ converges

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 $n\to \infty$.

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.