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