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This question came up after going through the literature on sequences and sums. Under what conditions can we obtain this $1/n$ bound ?

Thanks in advance !

user385459
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2 Answers2

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It does not follow. Consider $$ a_n=\cases{\frac1n& if $n$ is a perfect square\\0& otherwise} $$which has finite sum, but is not $o(1/n)$.

An assumption like $a_n$ being a monotonic sequence is sufficient for $o(1/n)$, but I don't know any good (as in useful and / or interesting) sufficient and necessary conditions.

Arthur
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    To address your comment about $a_n>0$, you can change the "otherwise" line of my $a_n$ definition above to $1/n^2$ instead of $0$, and it still has finite sum without being $o(1/n)$. – Arthur Jul 09 '18 at 05:50
  • Arthur. What is o(1/n) in your example?Thanks, Peter – Peter Szilas Jul 09 '18 at 09:47
  • @PeterSzilas nothing is $o(1/n)$ in my example. That was the point. My example is a sequence with non-negative terms, and finite sum. Your question was whether it then follows that the sequence is $o(1/n)$, but my example isn't $o(1/n)$, so clearly it cannot follow. – Arthur Jul 09 '18 at 10:05
  • Arthur.Hi, thanks.I bug you a little more, if you don't mind.Both your subsequences go faster to 0 than 1/n, I.e o(1/n). Can you infer o(1/n) for the original a_n by considering the slower subsequence a_k =1/k^2 .I seem to miss something.Thanks again.Greetings. – Peter Szilas Jul 09 '18 at 10:17
  • @PeterSzilas $o(1/n)$ doesn't care what most of the terms do. It cares about what the "largest" terms do. And the "largest" terms in my sequence $a_n$ are $1/n$, and the sequence is therefore not $o(1/n)$. This argument is, of course, possible to write out rigorously, but that's the moral behind it. – Arthur Jul 09 '18 at 10:23
  • Arthur Thanks for your patience . – Peter Szilas Jul 09 '18 at 11:09
  • Thank you very much for these great examples! Meanwhile I have also come across this post https://math.stackexchange.com/questions/1135536/if-sum-n-0-infty-a-n-diverges-prove-that-sum-n-0-infty-fraca-n?rq=1 which discusses this deeper in the post. Maybe they should change their title given the history they cite! – user385459 Jul 12 '18 at 01:23
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You could have something like $$a_n=\begin{cases}1/k^2&n=k^3\\0&\text{otherwise}\end{cases}$$ Then $\sum a_n=\frac{\pi^2}6$, but $a_n\ne o(1/n)$; in fact, $a_n=\Omega(n^{-2/3})$.

The situation is different if the sequence $a_n$ is monotonic.

Hagen von Eitzen
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  • How does this change if $a_n$ is monotonic ? Assuming it is monotonic, I still cannot show that is it $o(1/n)$. I think i am missing something trivial! Thanks again for your answer – user385459 Aug 05 '18 at 16:30