real analysis - A divergent series, whose underlying sequence converges faster to $0$ than $n^{-1}$ and slower than $n^{-1-epsilon}$. for any $epsilon > 0$

Let $\left(a_n\right)_{n \in \mathbb N}$ be a monotone sequence of positive real numbers, such that $$\liminf_{n \to \infty} na_n = 0$$ Does it always follow that $$\sum_{n = 0}^\infty a_n < \infty$$
It is easy to show that $\sum_{n=0}^\infty a_n < \infty$ implies $\liminf_{n \to \infty} na_n = 0$, even if $(a_n)_{n \in \mathbb N}$ is not a monotone sequence. Moreover, one can easily come up with plenty of non-mononote sequences $(a_n)_{n \in \mathbb N}$, satisfying $\liminf_{n \to \infty} na_n = 0$, but $\sum_{n=0}^\infty a_n = \infty$. It is also easy to see that the existence of some $\epsilon > 0$, such that $\limsup_{n \to \infty} n^{1+\epsilon} a_n < \infty$ imples the convergence of $\sum_{n = 0}^\infty a_n$, again regardless of whether $(a_n)_{n \in \mathbb N}$ is monotone or not. However, I am stuck at this particular question. If there exists a counterexample, it should, more or less, be a sequence $(a_n)_{n \in \mathbb N}$ converging to $0$, but simultaneously slower than any sequence of the form $n^{-(1+\epsilon)}$ for all $\epsilon > 0$ and faster than than $n^{-1}$.

Answer

You can take $\frac 1{n \log(n)}$, for example. The sum diverges. You can also take $\frac 1{n(\log (n))^2}$, for which the sum converges.