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.
Consider
$$1+{1\over2}+{1\over2}+{1\over2}+{1\over2}+{1\over3}+{1\over3}+\cdots$$
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 $k\ge m$.
