2

I tend to think this is true. If it isn't then:

  1. If $\lim_{n\rightarrow \infty} na_n = L \neq 0$, then, we get: $L-\varepsilon \leq na_n \rightarrow \frac{L-\varepsilon}{n}\leq a_n$ And because $\sum_{n = 1}^{\infty} \frac{L-\varepsilon}{n}$ diverges, so does the original sum.
  2. If $\lim_{n\rightarrow \infty} na_n = \infty$, we can disprove in a similar way.
  3. If $\lim_{n\rightarrow \infty} na_n$ diverges. Here I got stuck, and also can't find a counter-example to the original claim.
sadcat_1
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  • Note that a_n is non-negative, so it's not a duplicate – sadcat_1 Feb 15 '22 at 22:32
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    Suppose $a_n = \frac1{n}$ when $n$ is a square and $0$ otherwise. – Henry Feb 15 '22 at 22:39
  • For (3) take a subsequence – Jake Freeman Feb 15 '22 at 22:42
  • @Henry You and currently four others clearly think that that's a good solution. Why did you put it in the comment section? – Arthur Feb 15 '22 at 23:02
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    @Arthur because some people at this site target certain types of answers and the answerers and travel in packs to do so. They find it harder to target comments. Using comments also shows that this issue is not about reputation – Henry Feb 15 '22 at 23:40
  • @Henry If you care about demonstrating that you don't care about reputation, you can use the Community Wiki checkmark. And some people target certain answers because answers to certain questions are decided to be unwanted. And you decide that you want to go behind the back of the official code of conduct here on the site and answer a question that maybe shouldn't be answered by putting an answer in a comment box that explicitly tells you not to put answers in it? Seems a little fishy to me. – Arthur Feb 15 '22 at 23:44
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    @Arthur - not a debate to have here. Unfortunately I am not willing to go to meta either because it is inhabited by unfriendly people – Henry Feb 15 '22 at 23:48

1 Answers1

4

Here's why it's obviously false. Start with any convergent series $\sum a_n$ with $a_n>0$. Let $(b_n)$ be the same sequence with zeroes inserted between terms: $$a_1,0,0,\dots,0,a_2,0,0,0,\dots,a_3,0,\dots.$$

It's clear that $\sum b_n$ converges, but if we add enough zeroes at each stage we will have $\limsup nb_n=\infty$.

Edit. It's been stated that this is not so, for example if $a_n=1/n^2$. Let's work that out in detail to clarify whatever it is that needs to be clarified: If we start with that sequence and insert enough zeroes we get $(b_n)$, where $b_{2^n}=a_n$, $b_k=0$ if $k\ne 2^n$. Hence $2^nb_{2^n}=2^n/n^2\to\infty$ (so $\limsup kb_k=\infty$).

David C. Ullrich
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