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Let $(a_n)$ be a sequence of positive non-increasing real numbers such that $$ \sum_{n \geq 1} a_n < \infty. $$ Then is it true that there exists $C>0$ such that $a_n n \log(n) \leq C$ for every $n \geq 1$? I know that $a_n n \to 0$ and also that $a_n n \log n \not \to 0$ as reported here. But for all the examples the sequence $a_n n \log (n)$ was still bounded. Furthermore all the counterexamples I tried failed to not be decreasing, or to not be different from zero, or to not be a summable sequence. But I am not either able to find a proof of this statement.

cbx
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    +1 for interesting question. – Adam Rubinson Dec 13 '23 at 12:24
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    Try to modify the example in https://math.stackexchange.com/a/2711750/42969, e.g. $a_n=\frac{\log(k)}{k^2 e^{k^2}}$ for $e^{(k-1)^2}< n \le e^{k^2}$. – Martin R Dec 13 '23 at 12:52
  • I woud try: If $n$ is a power of $2$, then $a_n = 1/n$, otherwise $a_n =2^{-n}$. Then $\sum a_n$ converges. But, along the subsequence of powers of $2$, we have $a_n n \log(n) = \log(n) \to \infty$. – GEdgar Dec 13 '23 at 18:59

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Adapting the counter-example in the body of my previous question here, the following is a counter-example (which is also related to Martin's comment):

$$\text{ For each } k\in\mathbb{N},\ \text{ let } a_n = \frac{ k }{ \left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right) } \text{ for all }\ 2^{2^k} < n\leq 2^{2^{k+1}} = \left(2^{2^{k}}\right)^2 .$$

Then,

$$\sum_{n=5}^{\infty} a_n = \sum_{k=1}^{\infty} \left( \left( \left( 2^{2^k}\right)^2 - 2^{2^k} \right) \frac{ k }{ \left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right)} \right)$$

$$ \leq \sum_{k=1}^{\infty} \left( \require{cancel}\cancel{\left( 2^{2^k}\right)^2} \frac{ k }{ \require{cancel}\cancel{\left(2^{2^{k}}\right)^2} \log\left( \left(2^{2^{k}}\right)^2 \right)} \right) = \sum_{k=1}^{\infty} \frac{ k }{ 2\log( 2 ) \cdot 2^k},$$

which we know converges.

Furthermore, whenever $n=\left(2^{2^{k}}\right)^2, k\in\mathbb{N},\ a_n = a_{\left(2^{2^{k}}\right)^2} = \frac{ k }{ \left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right)}, $ and so for these values of $n,$

$$ a_n n\log n = \frac{ k }{ \require{cancel}\cancel{\left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right)} } \cdot \require{cancel}\cancel{\left(2^{2^{k}}\right)^2 \log\left( \left(2^{2^{k}}\right)^2 \right)} = k\to\infty \text{ as } n\to\infty. $$

Adam Rubinson
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