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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.

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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.

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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 $k\ge m$.

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