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

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

• What standard calculus approaches did you try, and can you show your work from one such approach? – Namaste Nov 14 '18 at 19:19
• @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. – Makina Nov 14 '18 at 19:44
• I think that since Makina, has included what they tried in the comments, the close votes should be retracted... – Rustyn Nov 14 '18 at 20:00
• This is related to Flint Hills series. The convergence of such sequence depends on the irrationality measure of $\pi$. – i707107 Nov 14 '18 at 20:08
• A related question on MO: mathoverflow.net/q/24579 – user587192 Nov 14 '18 at 20:11

## 2 Answers

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.

• 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... – Rustyn Nov 14 '18 at 20:53
• (+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. – T. Bongers Nov 14 '18 at 20:53
• Does the exponent of $3$ matter? For other powers the sequence still has "poles" at multiples of $\pi$. – M. Nestor Nov 14 '18 at 20:53
• @M.Nestor A sufficiently large exponent there would force convergence. – T. Bongers Nov 14 '18 at 20:54
• @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.) – Micah Nov 14 '18 at 20:56

Thought I would include a visualization for interested parties: • Those outliers look like spurious data. Except, of course, they aren't. =) – user21820 Nov 15 '18 at 14:25
• What's the scale on this plot? Does it get as far as $n=355$? – Micah Nov 16 '18 at 19:41
• If I remember it correctly it's $n=1$ to $n=800$ – Rustyn Nov 16 '18 at 19:43