# Is $\lim\limits_{n\to\infty}\Bigl(\frac{2+\sin n}{3}\Bigr)^n =0$; $n \in \mathbb{N}$?

$$\lim_{n\to\infty}\Bigl(\frac{2+\sin n}{3}\Bigr)^n$$ where $$n$$ is a natural number.

The problem is if the numerator is less than 3. I think it is because $$\sin n$$ is never $$1$$ otherwise $$\pi$$ would be rational. But i am not sure if the limit exists because $$\sin$$ oscillates. How to solve this ?

• @ShubhamJohri those are not subsequences.
– user658409
Commented Aug 28, 2019 at 17:18
• You need to put $n$ an integer in your question.
– Mike
Commented Aug 28, 2019 at 17:49
• @Mike It is convention that $n\in\mathbb N$. Everybody knows this. So no need to do this. Commented Aug 28, 2019 at 18:17
• I made a graph. Here is the link Commented Aug 28, 2019 at 19:35
• For $n=573204$ the argument of the limit is very close to $1$, and there are infinite similar cases, granting that the limit does not exist. Commented Aug 28, 2019 at 22:35

The value of the limit, if existing, depends on how close to $$1$$ (and how often) $$\sin n$$ can be. Of course $$\sin n\approx 1$$ iff $$n$$ is close to $$\frac{\pi}{2}$$ plus an integer multiple of $$2\pi$$. Let us consider the convergents $$\frac{p_n}{q_n}$$ of the continued fraction of $$\pi\not\in\mathbb{Q}$$: they all fulfill $$\left|\pi q_n-p_n\right| \leq \frac{1}{q_n}.$$ Let us assume $$\color{red}{p_n\equiv 0\pmod{2}\text{ and }q_n\equiv 1\pmod{4}}$$, i.e. $$p_n=2a$$ and $$q_n=4b+1$$. Then

$$\left|\frac{\pi}{2}(4b+1)-a\right| \leq \frac{1}{8b}$$ implies $$|1-\sin a|\leq\frac{1}{a}$$, so $$\left(\frac{2+\sin a}{3}\right)^a$$ is at least as large as $$\exp\left(-\frac{1-\varepsilon}{3}\right)$$.

For the same reason $$\left(\frac{2+\sin p_n}{3}\right)^{p_n}$$ is extremely close to zero, hence the limit does not exist, provided that the red constraint holds infinitely often.

• OP I accidently downvoted your answer. I just upvoted. My apologies
– Mike
Commented Aug 28, 2019 at 17:48
• Why is $1-\sin a\le\tfrac 1a$? When I use your line below the red part, I get $1-\sin a\le\tfrac 1{8b}$. Commented Aug 28, 2019 at 18:00
• @JackD'Aurizio No, $\frac a{4b+1}\approx\frac\pi 2$, hence $\frac a{4b}\approx\frac\pi 2$ and so $\frac{b}a\approx\frac 1{2\pi}$. But that also does it, thanks. Commented Aug 28, 2019 at 18:23
• @amsmath: of course you are right. Still $\frac{8}{2\pi}\gg 1$, so the inequalities above are fine. Commented Aug 28, 2019 at 18:26
• @Milan: $\sin$ is a $1$-Lipschitz function, so if $a$ and $b$ differ by less than $d$, then $\sin(a)$ and $\sin(b)$ also differ by less than $d$. The final part just follows from $\left(1+\frac{C}{n}\right)^n\approx e^C$ for any $n$ large enough. Commented Aug 29, 2019 at 18:40