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

Question

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

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

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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

Answer 1

Thought I would include a visualization for interested parties:

Answer 2

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 $C \approx 7.6$.)

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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.

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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 $\mu(\pi)>4$.

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