If $\,a_n\searrow 0\,$ and $\,\sum_{n=1}^\infty a_n<\infty,\,$ does this imply that $\,n\log n\, a_n\to 0$?

Question

A quite elegant and classic exercise of Calculus (in infinite series) is the following:

If the non-negative sequence $\{a_n\}$ is decreasing and $\sum_{n=1}^\infty a_n<\infty$, then $na_n\to 0$.

To show this observe that, if $\,\sum_{n\ge n_0}a_n<\varepsilon/2$, then for for $n\ge 2n_0+1$, $$\frac{\varepsilon}{2}>a_{\lfloor n/2\rfloor}+\cdots+a_n\ge \frac{1}{2}na_n\ge 0.$$

This, in a sense, is related to the fact that $\sum\frac{1}{n}=\infty$. To go one step further, since $\sum\frac{1}{n\log n}=\infty$, can we obtain, with the same assumptions on $\{a_n\}$, that $\,n\log n\, a_n\to 0$?

This conjecture holds with the additional assumption that $b_n=na_n$ is also decreasing. To see this, let $n_0\in\mathbb N$, such that $\sum_{n\ge n_0}a_n<\varepsilon$. Then for $n\ge n_0^2$, we have $$ \varepsilon>\sum_{\sqrt{n}\le k\le n}a_n\ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor}\sum_{k=\lfloor2^{\ell-1}\log_2\sqrt{n}\rfloor+1}^{ \lfloor2^\ell\log_2\sqrt{n}\rfloor}a_k\ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor} (2^{\ell-1}\log_2\sqrt{n}-1)a_{\lfloor2^\ell\log_2\sqrt{n}\rfloor} \\ \ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor} \Big(\frac{1}{2}{\lfloor2^\ell\log_2\sqrt{n}\rfloor}-1\Big) a_{{\lfloor2^\ell\log_2\sqrt{n}\rfloor}}\ge \sum_{\ell=1}^{\lfloor\log_2 \sqrt{n}\rfloor}\Big(\frac{1}{2}n-1\Big)a_n\ge (\log_2 n-1)\Big(\frac{1}{2}n-1\Big)a_n. $$ Hence, $(\log_2 n-1)\big(\frac{1}{2}n-1\big)a_n\to 0$, which implies that $\,\,n\log n\,a_n\to 0$.

One step further: If $a_n>0$, $\sum_{n=1^\infty}a_n<0$ $a_n$, and the sequences $na_n$ and $n\log n a_n$ are decreasing, then $$ n\log n \log\log n\,a_n\to 0. $$

Answer 1

A counterexample is given by $$ a_n=\frac{1}{k^2 e^{k^2}} $$ when $e^{(k-1)^2}\leq n< e^{k^2}$. Then the number of terms equal to $1/(k^2 e^{k^2})$ is bounded above by $e^{k^2}$, so $\sum_n a_n\leq \sum_k 1/k^2<\infty$, but when $n$ is slightly less than $e^{k^2}$, $$ n\log n\,a_n\approx1, $$ so $n\log n\, a_n\not\to 0$.

Answer 2

Actually, there is nothing specific with $\log n$: the best we can say is that $n\cdot a_n\to 0$.

Let $\left(R_n\right)_{n\geqslant 1}$ be a sequence of non-negative numbers with goes to infinity. Then there exists a non-decreasing sequence of non-negative numbers $\left(a_n\right)_{n\geqslant 1}$such that $\sum_{n=1}^{+\infty}a_n$ is finite but $\left(nR_na_n\right)_{n\geqslant 1}$ does not converge to $0$.

Indeed, let $\left(N_k\right)_{k\geqslant 1}$ be an increasing sequence of integers such that $R_{N_{k+1}} \geqslant k^2$ for all $k\geqslant 1$. If $N_k\lt n\leqslant N_{k+1}$, let $$a_n:=\frac 1{k^2N_{k+1}}.$$ Then $$ \sum_{k=1}^{+\infty}\sum_{n=N_k+1}^{N_{k+1}}a_n=\sum_{k=1}^{+\infty}\sum_{n=N_k+1}^{N_{k+1}}\frac 1{k^2N_{k+1}}= \sum_{k=1}^{+\infty}\frac{N_{k+1}-N_k}{k^2N_{k+1}}\leqslant \sum_{k=1}^{+\infty}\frac{1}{k^2}<+\infty $$ and for all $k\geqslant 1$, $$ N_{k+1}R_{N_{k+1}}a_{N_{k+1}}=\frac 1{k^2}R_{N_{k+1}}\geqslant 1. $$

If we want a strictly decreasing sequence, add to $a_n$ a term $b_n$ which decreases to $0$ and such that $\sum_{n\geqslant 1}nR_nb_n$ is finite.