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Does the sequence $$a(n) = \frac{1}{n^3\sin(n)}$$ converge ?

I tried all possible standard calculus approaches but to no avail ...

edit:

I tried using the root theorem and the limit of the $\frac{a_{n+1}}{a_{n}}$ which kinda got me nowhere ... Then I followed it with trying to prove that $n^3\cdot \sin(n)$ has no lower bound $K > 0$ by checking the behavior of the function $|n^3\cdot \sin(n)|$ and concluding that at some point the integer value of $n$ will bring me the value of function, which will be between $0$ and $K$, but I failed to give a rigorous proof of that conclusion

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    $\begingroup$ What standard calculus approaches did you try, and can you show your work from one such approach? $\endgroup$
    – amWhy
    Commented Nov 14, 2018 at 19:19
  • $\begingroup$ @amWhy I tried using the root theorem, which gave me 1 in the limit, and the limit of the $\frac{a_{n+1}}{a_{n}}$ which kinda got me nowhere either ... Then I followed it with trying to prove that n3∗sin(n) has no lower bound K >0 by checking the behavior of the function |n3∗sin(n)| and concluding that at some point the integer value of n will bring me the value of function, which will be between 0 and K, but I failed to give a rigorous proof of that conclusion. $\endgroup$
    – Makina
    Commented Nov 14, 2018 at 19:44
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    $\begingroup$ I think that since Makina, has included what they tried in the comments, the close votes should be retracted... $\endgroup$
    – Rustyn
    Commented Nov 14, 2018 at 20:00
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    $\begingroup$ This is related to Flint Hills series. The convergence of such sequence depends on the irrationality measure of $\pi$. $\endgroup$ Commented Nov 14, 2018 at 20:08
  • $\begingroup$ A related question on MO: mathoverflow.net/q/24579 $\endgroup$
    – user587192
    Commented Nov 14, 2018 at 20:11

2 Answers 2

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The answer to this question depends on the irrationality measure $\mu(\pi)$ of $\pi$, in a way which means it is unsolved. (The current state of the art is that $2 \leq \mu(\pi) \leq C$, where $C \approx 7.6$.)


Suppose that $\mu(\pi)>4$. Then there exist infinitely many pairs of integers $(p,q)$ such that

$$\left|\pi - \frac{p}{q}\right|<\frac{1}{q^4}$$

For such a $p$, $|\sin p|=|\sin(p-q\pi)|<|q\pi - p|<\frac{1}{q^3}$ and so $$ \left|\frac{1}{p^3\sin p}\right|>\frac{q^3}{p^3}>\frac{1}{27} $$ (as $\frac{p}{q}$ closely approximates $\pi$, so in particular it will be greater than $3$). Since the sequence can only converge to zero, this is enough to show that it diverges.


On the other hand, suppose the sequence diverges. Then there is some constant $C$ and subsequence $(p_n)$ such that $$\left|\frac{1}{(p_n)^3\sin p_n}\right|>C$$ for all $n$. Choose $q_n$ so that $|p_n-\pi q_n|<\frac{\pi}{2}$. Then we have $$ |\pi q_n-p_n|<\frac{\pi}{2}|\sin(p_n-\pi q_n)|=\frac{\pi}{2}|\sin p_n|<\frac{1}{C (p_n)^3} $$ and so $$ \left|\pi-\frac{p_n}{q_n}\right|<\frac{1}{C(p_n)^3q_n}<\frac{1}{27C(q_n)^4} $$

for infinitely many $p_n,q_n$. This is enough to imply that $\mu(\pi)>4$.


So, in summary, finding whether the sequence converges essentially boils down to comparing $\mu(\pi)$ to $4$: a wildly unsolved problem.

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  • $\begingroup$ Many thanks for this answer, I think it's the most complete yet. Somehow I knew that this question wasn't straightforward as one might expect... $\endgroup$
    – Rustyn
    Commented Nov 14, 2018 at 20:53
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    $\begingroup$ (+1) for a nice answer that actually addresses how hard this is. For interested readers, there is this note that bounds the irrationality measure of $\pi$ in a fairly elementary way, although it's not the best known result anymore. $\endgroup$
    – user296602
    Commented Nov 14, 2018 at 20:53
  • $\begingroup$ Does the exponent of $3$ matter? For other powers the sequence still has "poles" at multiples of $\pi$. $\endgroup$
    – user519413
    Commented Nov 14, 2018 at 20:53
  • $\begingroup$ @M.Nestor A sufficiently large exponent there would force convergence. $\endgroup$
    – user296602
    Commented Nov 14, 2018 at 20:54
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    $\begingroup$ @M.Nestor: If the exponent is larger than $\mu(\pi)-1$, the series converges. (So $\frac{1}{n^7 \sin n}$ is definitely convergent, but we're not sure about $\frac{1}{n^6 \sin n}$ yet.) $\endgroup$
    – Micah
    Commented Nov 14, 2018 at 20:56
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Thought I would include a visualization for interested parties: Sequence

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    $\begingroup$ Those outliers look like spurious data. Except, of course, they aren't. =) $\endgroup$
    – user21820
    Commented Nov 15, 2018 at 14:25
  • $\begingroup$ What's the scale on this plot? Does it get as far as $n=355$? $\endgroup$
    – Micah
    Commented Nov 16, 2018 at 19:41
  • $\begingroup$ If I remember it correctly it's $n=1$ to $n=800$ $\endgroup$
    – Rustyn
    Commented Nov 16, 2018 at 19:43

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