Prove or disprove convergence $\sum (a_n)^m$

Problem :

Given sequence $$\left\{a_n\right\}$$ : $$\forall n\in\mathbb{N}, a_n > 0, \quad \lim_{n\to\infty}a_n=0$$

Does there always exist some positive real number $$m$$ which makes series $$\sum_{n=1}^\infty (a_n)^m < \infty$$ converge?

I think this is false and I guess there is counterexample but I can't construct it.

I tried to make $$a_n < \frac{1}{n}$$ and take $$m>1$$ but I think this approach isn't good.

Thanks for any help or hints.

Consider $$a_{n}=\dfrac{1}{\log n}$$ for $$n\geq 2$$, then for any $$m>0$$, there is some constant $$C_{m}>0$$ such that $$\log n\leq C_{m}n^{1/m}$$, and hence $$a_{n}^{m}\geq C_{m}^{-1}n^{-1}$$, the rest is clear.
$$1+{1\over2}+{1\over2}+{1\over2}+{1\over2}+{1\over3}+{1\over3}+\cdots$$
where each fraction $$1/k$$ appears $$k^k$$ times. Then, on grouping terms, we have
$$\sum(a_n)^m=\sum_{k=1}^\infty k^{k-m}=\infty$$
since $$k^{k-m}\ge1$$ for all $$k\ge m$$.