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The limit is: $$\lim_{x\to\infty}\frac1{x^3\sin^2(x)}$$ Is there a way to prove this without using expansions?

J. M. ain't a mathematician
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    The limit fails to exist since for $x=n\pi$, $\sin(n\pi)=0$ and the denominator is undefined. – Mark Viola Mar 23 '17 at 05:01
  • @Dr.MV matlab and maple give me the evaluation of 0 –  Mar 23 '17 at 05:03
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    @Denis In that case, both Matlab and Maple are wrong. The limit does not exist for the reason that Dr. MV described. – Théophile Mar 23 '17 at 05:04
  • @Théophile so if I have $\sum_{n=1}^\infty \frac1{n^3\sin(n)^2}$ it would mean that it diverges? –  Mar 23 '17 at 05:06
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    @Denis Not necessarily. Here, $n$ is an integer. So, extra attention to that detail is required. – Mark Viola Mar 23 '17 at 05:09
  • @Dr.MV Can you explain in more detail? –  Mar 23 '17 at 05:10
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    See http://math.stackexchange.com/a/20609/148510. The convergence/divergence of that series is not known. – RRL Mar 23 '17 at 05:25
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    Now that you know that the convergence of the series is unknown, and that convergence of the limit over the reals is false, what are you still looking for? –  Mar 23 '17 at 06:28
  • @MikeMiller The limit with x in the natural numbers. I want to understand in more detail what Dr. MV stated in his second comment. Precisely: "$n$ is an integer so extra attention to that detail is required" –  Mar 23 '17 at 06:34
  • The graph of $1/(x^3\sin^2 x)$ has infinitely many singularities at which its value blows up to $+\infty$, so even if you restrict the domain to not have those singularities the function will not converge as $x\to\infty$. This is not enough to conclude it fails to converge as $n\to\infty$ in the naturals, because you don't know how close the naturals tend to get to the singularities asymptotically. – anon Mar 23 '17 at 06:38
  • @Denis Typically, in the absence of other context, $x$ is used for real numbers and $n$ for natural numbers. So, now that it's established that you want to restrict the variable to natural numbers, you could write the question thus: $$\lim_{n\to\infty}\frac1{n^3\sin^2(n)}$$ – Théophile Mar 23 '17 at 14:30

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Nobody knows whether that limit converges over the naturals, but expects the answer is "yes, to zero".

The notation and solution more or less come from here. Consider the $n$ such that $|n^3\sin^2(n)| \geq c$. Write $q_n$ the natural minimizing $|q_n\pi - n|$. Then $|n^3 \sin^2(n)| = |n^3 \sin^2(q_n\pi-n)|$.

That this is $\geq c$, then $\sin^2(q_n\pi-n) \geq cn^{-3}$, and (sorry!) because the power series for $\sin^2$ is alternating (or by some elementary geometry), we see that $(q_n\pi - n)^2 \geq \sin^2(q_n\pi - n) \geq cn^{-3}$, and therefore $|\pi-n/q_n| \geq cn^{-3/2}q_n^{-1}$. For this to be nearly $\pi$, $q_n \simeq n/\pi$, so (perhaps changing the constant) we have that for relevant $n$, $cn^{-3} \geq Cq_n^{-3},$ so $|\pi - n/q_n| \geq Cq_n^{-5/2}$. If there are infinitely many such $n$, we see that the irrationality exponent of $\pi$ is less than or equal to $5/2$. As mentioned in this answer, this is open, and if it was strictly less than $5/2$ the summation of your series converges, too.

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