Supose $\displaystyle(b_n)_{n \in \mathbb{N}}$ is a sequence of positive real numbers that
$$\displaystyle\sum_{n \in \mathbb{N}}(b_n)^{2} <\infty.$$
Does exists some $\epsilon>0$ such that $\displaystyle\sum_{n \in \mathbb{N}}(b_n)^{2- \epsilon} <\infty$ also converges?
The number $2$ isn't special here, it could be any positive number, one more general statement could be that $\displaystyle\sum_{n \in \mathbb{N}}(b_n)^{t} <\infty$ only converges for $t$ in open intervals of the line. I now that if this series converges to some point $t^\ast$ it must converges to all points $t > t^\ast$, because $\ell^p \subset \ell^q$ if $q<p$, at least if $p,q>1$.
I don't know if this statments are true or if there is some conter-example, a think that this kind of result is related to general Dirichlet series and Abscissa of convergence of that kind of series, but I don't now much about this topics.
Recall that for monotone nonincreasing sequences $a_1\ge a_2\ge \dots \ge 0$, the series $\sum a_n$ converges if and only if the series $$ \sum_k 2^ka_{2^k} $$ converges. This is the Cauchy condensation test. You can check using this test that the series $$ \sum_{n=3}^\infty \frac{1}{n(\log n)^2} $$ converges, whereas for any $\epsilon>0$, $$ \frac{n^\epsilon(\log n)^{2\epsilon}}{n(\log n)^2}\ge \frac 1{n^{1-\epsilon}(\log n)^2}, $$ and the series $\sum\frac 1{n^{1-\epsilon}(\log n)^2}$ diverges, since $$ \sum_k \frac {2^k}{2^{k(1-\epsilon)}(\log 2^k)^2} = \sum_k \frac{2^{k\epsilon}}{k^2}\quad\text{diverges.} $$
On the other hand, if $\sum a_n$ is a convergent series of positive terms, then $\sum a_n^{1+\epsilon}$ converges, because $a_n^{1+\epsilon}\le a_n$ eventually if $\epsilon>0$.