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Let $(a_n)$ be a sequence of real positive numbers such that $\sum a_n$ is a convergent series. What can we say about the series $\sum a_n^{\frac{n-1}{n}}$? Show that it is convergent or find a counterexample.

I'm trying to find a counterexample but maybe is true that the series is convergent.

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  • $\begingroup$ It looks interesting. Write $b_n := a_n^{(n-1)/n}$. If $$\limsup \frac{b_n}{a_n} < +\infty,\tag{1}$$ then convergence of $\sum b_n$ follows from convergence of $\sum a_n$. And the only counterexamples I have found to (1) have $a_n$ going so rapidly to zero that $b_n$ still goes rapidly to zero and $\sum b_n$ converges. $\endgroup$ – GEdgar Jul 31 '18 at 11:41
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$\sum_n a_n=\sum_{n \in I} a_n +\sum_{n \in J} a_n$ where $I=\{n:a_n <2^{-n}\}$ and $J=\{n:a_n \geq 2^{-n}\}$. If $n \in I$ then $a_n^{\frac {n-1} n}<2^{-{n+1}}$ so the first sum is convergent. If $n \in J$ then $a_n^{-1/n}\leq e^{-\frac 1 n \log \frac 1 {2^{n}}}=2$ so $a_n^{\frac {n-1} n} \leq 2a_n$ which makes the second series also convergent.

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