How to prove this sequence tends to zero

Question

Suppose $a_n>0$,$\sum a_n$ converges, $\{na_n\}_{\Bbb N}$ is monotonic, prove $$\lim na_n\ln n=0$$

My attempt so far has shown that $\{na_n\}$ is decresing: otherwise, $na_n\ge a_1\implies a_n\ge\frac{a_1}{n}\implies \sum a_n=+\infty$ which contradicts the convergence of $\sum a_n$.

My friend has got a bit further: $(n+1)a_{n+1}< na_n\implies \frac{a_{n+1}}{a_n}<\frac{n}{n+1}$. He then let $$b_n:=\frac{a_n}{\frac{1}{n\ln n}}$$ and therefore $$\frac{b_{n+1}}{b_n}=\frac{a_{n+1}}{a_n}\cdot\frac{(n+1)\ln(n+1)}{n\ln n}<\frac{\ln(n+1)}{\ln n}$$ However he can't prove the monotony of $b_n$ because the RHS of the inequality is too weak. In fact it only shows $\limsup \frac{b_{n+1}}{b_n}\le 1$, which is not effective here. And I can't come up with a stronger version, either.

Can you help me? Any direct help or hint will be appreciated. Thanks!

@MarcoCantarini thank u. Have voted to close. – Vim Jun 26 '15 at 11:20 — Vim, Jun 26 '15 at 11:20

score 2 · Answer 1 · answered Jun 26 '15 at 06:30

2

I think it is a good idea to try expanding $\ln n \approx \sum_{k=1}^n \frac{1}{k}$.

This helps using the condition that for $m>n \Rightarrow \frac{a_n}{m} \ge \frac{a_m}{n}$.

answered Jun 26 '15 at 06:30

João Ramos

1,770

Thanks! Well, that can show $b_n$ is upper-bounded. But what about the monotony? – Vim Jun 26 '15 at 06:40
Well, actually it was only a hint :P – João Ramos Jun 26 '15 at 06:40
You can complete the proof by fixing $k_0$ and summing inverses only from this $k_0$ On. See what happens in this case. – João Ramos Jun 26 '15 at 06:41

score 2 · Answer 2 · answered Jun 26 '15 at 07:58

2

I haven't enough experience yet with series, so please let me know if the following reasoning is incorrect. I may suspect it is since I don't use all the information. In that case I'll delete the answer right away.

Assume $\lim\limits na_n\ln n\ne0,$ or doesn't even exist. Then there is at least a subsequence of $\{na_n\ln n\}$ bounded below by some positive $\varepsilon$, whence we can write, for a suitable $k$, $$\sum a_n>\sum_{n>k}\frac{\varepsilon}{n\ln n}=+\infty,$$ which contradicts the convergence of $ \sum a_n $.

answered Jun 26 '15 at 07:58

Vincenzo Oliva

7,249

Sorry I was just too quick to judge. Maybe you should first shortly justify the existence of such a $k$? – Vim Jun 26 '15 at 08:03
@Vim Correct me if I'm wrong, but I think this doesn't work. If we take $a_n>\frac{\varepsilon}{n\ln n}$ only when $n$ is a square, we get no contradiction about the convergence. – Vincenzo Oliva Jun 26 '15 at 08:22
@Vim I guess I'm right? – Vincenzo Oliva Jun 26 '15 at 13:29
would you care to elaborate? – Vim Jun 26 '15 at 13:36
@Vim What's wrong with the comment above? – Vincenzo Oliva Jun 26 '15 at 13:38
You said "it doesn't work"? – Vim Jun 26 '15 at 13:39
in fact. It's vital to justify that there exists a k such that for all integers greater than k we have $a_n>\epsilon/n\ln n$ – Vim Jun 26 '15 at 13:41
@Vim I meant that the approach in my answer probably doesn't work, because the subsequence may also be too sparse to yields a divergent series. – Vincenzo Oliva Jun 26 '15 at 13:43
Shouldn't $\sum\frac{1}{n\ln n}$ be $\sum\frac{1}{{n_k}\ln {n_k}}$? – Tony Ma Apr 14 '18 at 03:57

How to prove this sequence tends to zero

2 Answers2