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I am asking if this series
$$ \sum_{n=1}^\infty \frac{1}{n^2\sin n} $$ is convergent or divergent?

To "study" this I used the $n$th term test wich says: if the limit of $ \frac{1}{n^2\sin n}$ isn't zero or it doesn't exist, than the series has to be divergent. By following this, as the limit doesn't exist, the series has to be divergent, but there's a problem: series like $(-1)^n$ are NOT divergent. Let me explain. it depends of how you define a divergent series: I've seen that in some countries there's a distinction between divergent and "swinging" or irregular series like the example. I've also seen that in the US (for example on Wikipedia and many other books) that there is NO distinction between divergent and irregular series, and so every series of wich the limit IS NOT zero, than is divergent. I don't know so if the $n$th term test is valid or not. Sorry for my bad English.

  • That's an open problem. – Another User Oct 05 '22 at 13:42
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    Do you define swinging or irregular to be that the number of values that subsequences can approach is finite and more than one? – gist076923 Oct 05 '22 at 13:51
  • wolfram says that the series is DIVERGENT, by using the nth term test. Im not sure if the limit of An is indeterminate. If the series si divergent, than also the absolute one is divergent, than, referring to Alekseyev's work, the lower bound of the irrationality measure of pi has to be at 3. –  Oct 05 '22 at 13:51
  • @gst076923 yes i think. –  Oct 05 '22 at 13:53
  • another problem: if we follow the nth term test, and if we study series with An=1/(n^u(sin(n))) where u is a positive real number, and if we say that every u we put, then the limit of An is indeterminate, then always referring to Alekseyev's work, as this series are divergent for the nth term test, the lower bound of the irrationality measure of pi should be higher than 1+u. If we do this process for every u way and way bigger, than we find that the lower bound has to be +infinite and so pi is a liouville number, which is NOT possible and its like a wtf. –  Oct 05 '22 at 13:57
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    Your previous question to the same topic https://math.stackexchange.com/questions/4545715/exercise-about-convergence-divergence-of-a-series-sum-n-1-infty-frac1n got an answering comment linking to https://math.stackexchange.com/questions/2034910/why-this-sum-limits-n-1-infty-frac1n-sin-n-is-not-harmonic-series-and-i and the paper https://arxiv.org/abs/1104.5100. Please react to that, refine your question, ... Do not open duplicates. – Lutz Lehmann Oct 05 '22 at 14:03
  • @LutzLehmann i was wrong with the previous question, and i changed it. That's a new question now. –  Oct 05 '22 at 14:07
  • @dave Yes, but the paper I cited for you is still relevant. – Gary Oct 05 '22 at 14:08
  • @Gary so that's an open problem. I know, but the problem is if the limit of An is zero or doesnt exist, that's what im asking. –  Oct 05 '22 at 14:11
  • What you mean is a generalized limit, a sequence of functionals $\theta_n$ on the sequence space that gives the limit for convergent sequences, $\lim_{n\to\infty}\theta_n(a)=\lim_{n\to\infty}a_n$, but where the left side can converge for some divergent $a$. The Cèsaro mean is one example, $\theta_n(a)=\frac1n(a_1+a_2+...+a_n)$. But such an averaging process still does not converge for all sequences, and different averaging processes can give different results for the same divergent sequence. – Lutz Lehmann Oct 05 '22 at 14:13
  • Dave, please read the paper that was linked: https://arxiv.org/pdf/1104.5100.pdf. It contains a relatively elementary discussion, provided you have some familiarity with the notion of Diophantine approximation. It tells you exactly whether or not we know for sure if $n^{-2}\csc(n)\to0$. Indeed, in their notation, $1+u/v=3$. It is not known whether or not $\mu(\pi)<3$, $\mu(\pi)=3$ or $\mu(\pi)>3$. So we do not know if this sequence converges to zero – FShrike Oct 05 '22 at 14:14
  • @FShrike I've already seen Alekseyev's work and many other. –  Oct 05 '22 at 14:16
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    @dave In that case the specific form $1/(n^2\sin n)$ is not really relevant. You could have asked the same question with $(-1)^n$. The general result is that if a series is convergent then its terms must tend to $0$. Equivalently, if the terms do not tend to $0$, the series is not convergent. – Gary Oct 05 '22 at 14:16
  • @Gary ok, thank you. I wanted to know why wolfram says that this series is divergent, and so also if the limit is equal to zero. does anyone know this? –  Oct 05 '22 at 14:18
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    @dave The opposite is not true: take $\sum 1/n$. The terms tend to $0$, but the series is still divergent. – Gary Oct 05 '22 at 14:19
  • @Gary yes i know, but if the limit is NOT zero, than the series has to be divergent, also if the limit doesnt exist/is indeterminate. so, in this case, is the limit indeterminate or we dont know? if so, what about the limit of n^2 sin(n)?? –  Oct 05 '22 at 14:22
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    I have already answered it above: "The general result is that if a series is convergent then its terms must tend to $0$. Equivalently, if the terms do not tend to $0$, the series is not convergent." – Gary Oct 05 '22 at 14:23
  • @Gary ok, i know, but in this specific case which is the result of the limit? this is what im asking right now. –  Oct 05 '22 at 14:24
  • FShrike explained to you above: we do not know. – Gary Oct 05 '22 at 14:27
  • @Gary ok thank you. that's all. –  Oct 05 '22 at 14:27

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