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$\sum a_n^3$ converges implies that $\sum \frac{a_n}{n}$ converges?

I guess it is wrong, since we do not assume $a_n\geq 0$. But how to find an example?

Clearly, $a_n$ should be sign-alternating, but such kind of $(-1)^n/n^{1/3}$ does not help.

xldd
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    Here's how to ask a good question. Follow these guidelines to get help in this forum. It's particularly important that you share your own work and thoughts on the problem to show that you have made a serious effort by yourself before asking for help, and you're not just trying to get others to solve it for you. – jjagmath Dec 17 '23 at 11:17
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    Why is this so obviously true if $a_n$ are all $>0?$ – Adam Rubinson Dec 17 '23 at 13:44
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    @AdamRubinson Hölder's inequality. – Gabriel Romon Dec 17 '23 at 13:53
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    I don't have much experience with Holder's inequality, but let me try. $\displaystyle\sum_{k=1}^{n} \frac{a_k}{k} = \sum_{k=1}^{n} a_k\cdot \frac{1}{k} \leq \left( \sum_{k=1}^{n} \left(a_k\right)^3 \right)^{1/3} \cdot \left( \sum_{k=1}^{n}\left(\frac{1}{k}\right)^{3/2} \right)^{2/3}\ $ and let $n\to\infty.\ $ Is this correct? – Adam Rubinson Dec 17 '23 at 15:17
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    I disagree with the closure. The last sentence of the post clearly shows the OP has given some non-trivial thought to the question. – Clement C. Dec 17 '23 at 23:24
  • What exactly in the link given by @jjagmath above on "how to ask a good question" was not done by the OP in the current question, that the people voting to close feel is critical? – Clement C. Dec 17 '23 at 23:29

1 Answers1

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Clearly, $a_n$ should be sign-alternating.

Not quite. It needs to change signs (otherwise, it's a positive series wlog), but alternating is not necessarily the best way to get a counter-example.

Try the sequence given by $$a_{3n-2} = a_{3n-1} = \frac{1}{(\log(n+1))^{1/3}}, \qquad a_{3n} = -\left(\frac{2}{\log(n+1)}\right)^{1/3} \tag{1}$$ for all $n\geq 1$. Since $a_{3n-2}^3 + a_{3n-1}^3 + a_{3n}^3 = 0$ for all $n$ and $\lim_{n\to\infty} a_n^3=0$, the series $\sum_n a_n^3$ converges.

Note however then $$ \sum_{n=1}^{3N} \frac{a_n}{n} = \sum_{n=1}^N \left(\frac{1}{(3n-2)\log^{1/3}(n+1)}+\frac{1}{(3n-1)\log^{1/3}(n+1)} - \frac{2^{1/3}}{3n\cdot \log^{1/3}(n+1)} \right) \tag{2} $$ and since $$ \frac{1}{(3n-2)\log^{1/3}(n+1)}+\frac{1}{(3n-1)\log^{1/3}(n+1)} - \frac{2^{1/3}}{3n\cdot \log^{1/3}(n+1)} \operatorname*{\sim}_{n\to\infty} \frac{2-2^{1/3}}{3n\log^{1/3} n} $$ by comparison (2) diverges.

Clement C.
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  • (I believe the derivation in the second part is correct, but worth double-checking my asymptotics. Also, try replacing the $\log(n+1)$ by something else, see how that goes: any function growing too fast there will not be a coutnerexample...) – Clement C. Dec 17 '23 at 11:24
  • You forgot the cubic root in the term $\frac{2^{1/3}}{3n\cdot \log(n+1)}$ – jjagmath Dec 17 '23 at 11:25
  • @jjagmath Thanks! should be better now. – Clement C. Dec 17 '23 at 11:27