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I'm confused how this series :$$\displaystyle \sum_{n=1}^\infty\frac1{n\sin n}$$ is a convergent series as wolfram show here and it's not harmonic series in the same time .

My question here is:

Why this :$$\displaystyle \sum_{n=1}^\infty\frac1{n\sin n}$$ is not harmonic series and is convergent ?

Martin Sleziak
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zeraoulia rafik
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    The harmonic series is $$\sum_{n=1}^{\infty} \frac{1}{n}$$ and it diverges! – Peter Nov 28 '16 at 21:33
  • This series is very different than the harmonic series, notice that the series of your question have infinite negative terms. –  Nov 28 '16 at 21:35
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    The series $$\sum_{n = 1}^\infty \frac{1}{n\sin n}$$ doesn't converge, its terms don't converge to $0$. – Daniel Fischer Nov 28 '16 at 21:37
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    @Peter I don't know, when I open the link, I get a plot of the first few partial sums and "computation time exceeded", it neither says it does nor it does not converge. – Daniel Fischer Nov 28 '16 at 21:39
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    I think that the claim that the terms do not converge to $0$ is wrong. Isn't $$\lim_{n\rightarrow\infty} \frac{1}{n\sin(n)}=0$$ ? – Peter Nov 28 '16 at 21:41
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    @DanielFischer WA is wrong a lot in some basic things, but I don't think this time it is that simple: the sequence does converge to zero, I think, as for any $;n\in\Bbb N;$ we can "squeeze" the general term between $;-\frac1n;$ and $;\frac1n;$ – DonAntonio Nov 28 '16 at 21:42
  • no the limit is not 0 because sin (n) hasn't a limt at infty – zeraoulia rafik Nov 28 '16 at 21:42
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    If $\frac{p}{q}$ is a convergent of $\pi$, then $$\biggl\lvert \pi - \frac{p}{q}\biggr\rvert < \frac{1}{q^2}.$$ Then $\lvert q\pi - p\rvert < \frac{1}{q}$ and hence $\lvert \sin p\rvert < \frac{1}{q} \approx \frac{\pi}{p}$. So for infinitely many $p$ (the numerators of the convergents of $\pi$ we have $\lvert p\sin p\rvert < 4$. – Daniel Fischer Nov 28 '16 at 21:42
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    @user51189 That doesn't necessarily matter as it is bounded in $;[-1,1];$ – DonAntonio Nov 28 '16 at 21:43
  • @Peter, the limit is not zero and dosn't exist because sin(n) is periodic, look here :https://www.wolframalpha.com/input/?i=lim%281%2F%28n+sin+n%29+,n+to+infty – zeraoulia rafik Nov 28 '16 at 21:50
  • @user51189 Consider that we only have the natural numbers here, not the real numbers. So, $\sin(n)\ne 0$. I think, Wolfram assumes real numbers. – Peter Nov 28 '16 at 21:52
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    @user51189 That still is wrong: the limit of $;\frac1n\sin n;$ is zero and still sine is periodic. The reasons you give are irrelevant for the existence of the limit and its value. – DonAntonio Nov 28 '16 at 21:52
  • ok, do u know why in ur case, because in numertor is correct because u can calculating it using variable change y=1/n and u will get siny/y y to zero , but in the latter case u can't since its multiplied by n in denomunator and no standard method exist for indterminate case to be vanish – zeraoulia rafik Nov 28 '16 at 21:55
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    @user51189 Since $\sin(n)$ is never $0$ for natural $n\ge 1$, we do not have an indeterminate case. So, the argument is not valid here. What might be the case : $\sin(n)$ is infinite many often so near to $0$ that the limit does not exist. – Peter Nov 28 '16 at 22:00
  • Could it be undecideable whether the sequence converges (or which limit it has ) ? – Peter Nov 28 '16 at 22:08
  • @Peter, the sequence is not bounded, so it cannot converge. Observe that $\sin n$ can be arbitrarily small frequently. We can try to prove that frequently exists some $n$ such that $(n \sin n)^{-1}>1$ –  Nov 28 '16 at 22:12
  • @Masacroso It must be small relative to $\frac{1}{n}$ , so we cannot conclude easily that the limit does not exist. $n$ makes a "battle" against $\ \sin(n)\ $ and it is not clear who "wins" in the long term. – Peter Nov 28 '16 at 22:16
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    @Peter We can easily show that $\liminf\limits_{n\to\infty} \lvert n\sin n\rvert \leqslant \frac{\pi}{\sqrt{5}}$. On the other hand, clearly $\limsup\limits_{n\to\infty} \lvert n\sin n\rvert = +\infty$. If the irrationality measure of $\pi$ is strictly greater than $2$, we have $\liminf\limits_{n\to\infty} \lvert n\sin n\rvert = 0$. We even have that if the irrationality measure of $\pi$ is exactly $2$ and the sequence of partial quotients is unbounded. It is very likely that it is unbounded. – Daniel Fischer Nov 28 '16 at 22:34
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    See the first answer here for an in depth discussion on the behavior of $ | n \sin n | $. – CodeLabMaster Nov 29 '16 at 02:11

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Hint: According to the answer by @J.R. given at this question the sequence \begin{align*} \left(\frac{1}{n\sin n}\right)_{n\geq 1} \end{align*} is not convergent. We conclude the series \begin{align*} \sum_{n=1}^\infty\frac{1}{n\sin n} \end{align*} is divergent.

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Markus Scheuer
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