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We have $a_{n}>0 $ and the series $ \sum a_{n} $ is convergent. Does it mean that $ a_{n}\times \ln{n}$ tends to 0, as n goes to infinity?

I couldn't find any counterexample, no sequence of the form $\frac{1}{n^{m}}$, where $m>1$, works. I also tried expressing $a_{n}$ as difference of two sums, but it didn't help.

Is there a counterexample?

Andrés E. Caicedo
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rulergraham
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2 Answers2

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Sure, there is such a sequence. Find a subsequence of $1/\log n$ whose series converges. Let $a_n=1/2^n$ unless $1/\log n$ is one of the terms of the subsequence, in which case $a_n=1/\log n$.

In this example, the sequence $a_n \log n$ does not converge, as there is a subsequence of it that is constantly equal to 1, while another subsequence converges to 0. This is no accident: If $b_n>0$ for all $n$ and $b_n\log n$ converges but not to 0, then $\sum b_n$ diverges because, if $b_n\log n\to r>0$, then $b_n>r/(2\log n)$ for all $n$ large enough.

Andrés E. Caicedo
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  • We are assuming in this counterexample that the limit doesn't exist, right? – user Dec 04 '19 at 15:17
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    Yes, in this example the limit does not exist. If the limit exists and is nonzero, the series must diverge, by basic comparison arguments. – Andrés E. Caicedo Dec 04 '19 at 15:18
  • That's could be a good point to add, even if basic. – user Dec 04 '19 at 15:21
  • You are right, thanks. – Andrés E. Caicedo Dec 04 '19 at 15:22
  • @user No, he's not assuming that the limit DNE! He's giving an example and showing that the limit does not exist in that example... – David C. Ullrich Dec 04 '19 at 16:17
  • @DavidC.Ullrich Yes indeed after my question this was clear from the reply by Andres. The limit can' t exist. Why are you repeating that? – user Dec 04 '19 at 16:25
  • @user I'm not repeating anything, I'm correcting something that was said. You said "We are assuming in this counterexample that the limit doesn't exist, right?". Andres said yes. In fact the answer is no. – David C. Ullrich Dec 04 '19 at 16:27
  • @DavidC.Ullrich Yes but then it was already clarified by Andres. You are repeating what Andres has already very well explained. Thanks anyway! Repetita juvant – user Dec 04 '19 at 16:28
  • @DavidC.Ullrich BTW could you take a look here? – user Dec 04 '19 at 16:30
  • @DavidC.Ullrich I would happy if you can also add an answer to that! I think it would be very useful. Thanks. Bye – user Dec 04 '19 at 17:49
  • @user I'm not sure whether you're suggesting I add an answer here or to that other question. I either case I don't have anything to say that hasn't been said. – David C. Ullrich Dec 04 '19 at 20:01
  • @DavidC.Ullrich In that other question, I think that also an aswer by you could be useful. Thanks anyway. Bye – user Dec 04 '19 at 20:04
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Consider $$ a_n=\frac1{n^2}+\frac{\left[n=2^{k^2}\land k\in\mathbb{Z}^+\right]}{\log_2(n)} $$ where $[\cdots]$ are Iverson Brackets. $$ \limsup_{n\to\infty}a_n\log(n)=\log(2) $$

robjohn
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