TITLE: Is $\lim\limits_{n\to\infty}\Bigl(\frac{2+\sin n}{3}\Bigr)^n =0 $; $n \in \mathbb{N}$?
QUESTION [7 upvotes]: $$\lim_{n\to\infty}\Bigl(\frac{2+\sin n}{3}\Bigr)^n $$ 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 ?

REPLY [4 votes]: 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.