Prove that if a positive series $\sum_{\nu =1}^\infty a_\nu$ is convergent and the sequence $(\nu a_\nu)_{\nu =1}^\infty$ is decreasing, then $\lim_{\nu\to\infty}(\nu\log\nu)a_\nu=0$.

I've been trying to prove this for days, but so far I've only managed to prove that if the limit exists, it is equal to 0.

Could someone give me a hint?


Suppose $na_n\log n$ does not tend to $0$. There exists $n_k$ increasing to $\infty $ such that $a_{{n_k}} n_k \log n_k \geq a$ for all $k$ for some $a>0$. Further, replacing $\{n_k\}$ be a subequence if necessary we may assume that $$ \log (n_{k+1}-1) -\log n_k >\frac 1 2 \log n_{k+1}$$ for all $k$. Now $$\sum_n a_n=\sum_k \sum_{n_k \leq n <n_{k+1}} \frac 1 n (na_n)\geq \sum_k \sum_{n=n_{k}}^{n_{k+1}-1} \frac 1 n (n_{k+1}a_{n_{k+1}})>\sum_k ({\log (n_{k+1}-1) -\log n_k}) (n_{k+1}a_{n_{k+1}}) $$. The general term of this series exceeds $\frac a 2$. This makes $\sum a_n$ divergent.

  • $\begingroup$ If you estimate the sum $\displaystyle \sum_{n=n_k}^{n_{k+1}-1} \frac{1}{n}$ by an integral the usual way, you get $\log(n_{k+1}) - \log(n_k)$ which looks nicer, although yours is also correct. $\endgroup$ – Adayah Jul 10 '18 at 8:49
  • $\begingroup$ I mentioned in my answer that this can be ensured by going to a subsequence. We can make $n_{k+1}$ 'much larger' than $n_k$ by going to a subsequence. $\endgroup$ – Kabo Murphy Jul 10 '18 at 10:06
  • $\begingroup$ @apanpapan3 Explicitly, $\frac {\log (n-1) -\log n_k} {\log n} \to 1$ so we can make this greater than $\frac 1 2$. $\endgroup$ – Kabo Murphy Jul 10 '18 at 10:09
  • $\begingroup$ @apanpapan3 What equivalence are you talking about? $\endgroup$ – Kabo Murphy Jul 10 '18 at 10:36

First notice that $(a_n)_n$ is also decreasing:

$$na_n \ge (n+1)a_{n+1} \implies a_n \ge \frac{n+1}{n} a_{n+1} \ge a_{n+1}$$

Since $\sum_{n=1}^\infty a_n$ converges, by Cauchy condensation test $\sum_{n=1}^\infty 2^na_{2^n}$ also converges.

Also, $(2^na_{2^n})_n$ is decreasing because it is a subsequence of $(na_n)_n$.

Now use this lemma:

Let $(x_n)_n$ be a decreasing sequence of positive numbers such that $\sum_{n=1}^\infty x_n$ converges. Then $\lim_{n\to\infty} nx_n = 0$.

We conclude that $\lim_{n\to\infty} n2^na_{2^n} = 0$.

By scaling, we can assume that $\log = \log_2$.

We have $\log n \le 2\lfloor \log n\rfloor$ and $n \ge 2^{\lfloor \log n\rfloor}$ so $na_n \le 2^{\lfloor \log n\rfloor} a_{2^{\lfloor \log n\rfloor}}$. Therefore

$$0 \le (n\log n) a_n \le \log n\cdot 2^{\lfloor \log n\rfloor} a_{2^{\lfloor \log n\rfloor}}\le 2\lfloor \log n\rfloor\cdot 2^{\lfloor \log n\rfloor} a_{2^{\lfloor \log n\rfloor}}$$

The right hand side converges to $0$ because it is a constant times a subsequence of $(n2^na_{2^n})_n$ with some repeated terms because $n\mapsto \lfloor \log n\rfloor$ is not strictly increasing.

The squeeze theorem gives $\lim_{n\to\infty} (n\log n) a_n = 0$.

  • $\begingroup$ 1. How do you know that $a_n$ is decreasing so that we can use the Cauchy condensation test? 2. The inequality $n a_n \leqslant 2^n a_{2^n}$ should go the other way. $\endgroup$ – Adayah Jul 10 '18 at 7:51
  • $\begingroup$ @Adayah You are right, of course. I tried to fix the issues, can you have a look now? $\endgroup$ – mechanodroid Jul 10 '18 at 23:26
  • $\begingroup$ $\mbox{}$Seems fine. $\endgroup$ – Adayah Jul 11 '18 at 15:15


The Mean Value Theorem guarantees that $$ \frac1n\ge\log(n+1)-\log(n)\tag1 $$ we have $$ \begin{align} \sum_{k=\left\lfloor\sqrt{n}\right\rfloor}^n\frac1k &\ge\log(n+1)-\log\left(\left\lfloor\sqrt{n}\right\rfloor\right)\\ &\ge\frac12\log(n)\tag2 \end{align} $$

The Answer

Suppose that $n a_n\searrow0$ and $\sum\limits_{n=0}^\infty a_n$ converges.

Assume that $\limsup\limits_{n\to\infty}n\log(n)\,a_n\gt0$; that is, $$ \exists\epsilon\gt0:\forall N\gt0,\exists\,n_N\ge N:n_N\log(n_N)\,a_{n_N}\ge\epsilon\tag3 $$ Since $na_n$ is decreasing, we have that for $k\le n_N$, $$ \begin{align} ka_k &\ge n_Na_{n_N}\\[6pt] &\ge\frac{\epsilon}{\log(n_N)}\tag4 \end{align} $$ Therefore, combining $(2)$ and $(4)$, we get $$ \begin{align} \sum_{k=\left\lfloor\sqrt{n_N}\right\rfloor}^{n_N}a_k &\ge\sum_{k=\left\lfloor\sqrt{n_N}\right\rfloor}^{n_N}\frac{\epsilon}{k\log(n_N)}\\ &\ge\frac\epsilon2\tag5 \end{align} $$ If we choose $N_{j+1}\ge\left(n_{N_j}+1\right)^2$, then $$ \left\lfloor\sqrt{n_{N_{j+1}}}\right\rfloor\ge n_{N_j}+1\tag6 $$ and so the ranges $\left[\left\lfloor\sqrt{n_{N_j}}\right\rfloor,n_{N_j}\right]$ are disjoint. Thus, by $(5)$, $$ \begin{align} \sum_{n=1}^\infty a_n &\ge\sum_{j=1}^\infty\sum_{k=\left\lfloor\sqrt{n_{N_j}}\right\rfloor}^{n_{N_j}}a_k\\ &\ge\sum_{j=1}^\infty\frac\epsilon2\tag7 \end{align} $$ Therefore, if $(3)$ is true, the sum in $(7)$ diverges. The contrapositive says that if the sum converges, then $$ \lim\limits_{n\to\infty}n\log(n)\,a_n=0\tag8 $$


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