Evaluate $ \lim_{n \to \infty} \sin^n\left(\frac{2\pi n}{3n+1}\right)$

Question

I'm trying to solve this limit rigorously

$ \lim_{n \to \infty} \sin^n\left(\frac{2\pi n}{3n+1}\right)$

I can see that the answer is $0$ as $$\lim_{n \to \infty} \sin\left(\frac{2\pi n}{3n+1}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \lt 1$$ But this solution doesn't seem rigorous and I'm unable to come up with a rigorous approach

Answer 1

For $n>2$, $$ \pi>\frac{2\pi}{3}>\frac{2\pi n}{3n+1}>\frac{2\pi n}{3n+n/2}=\frac{4\pi}{7}>\frac{\pi}{2}, $$ so for $n>2$, $$ \frac{\sqrt{3}}{2}=\sin\left(\frac{2\pi}{3}\right)<\sin\left(\frac{2\pi n}{3n+1}\right)<\sin\left(\frac{4\pi}{7}\right)<1. $$ Thus, $$ 0=\lim_{n\to\infty}\left(\frac{\sqrt{3}}{2}\right)^n\le \lim_{n\to\infty}\left(\sin\left(\frac{2\pi n}{3n+1}\right)\right)^n\le \lim_{n\to\infty}\left(\sin\left(\frac{4\pi}{7}\right)\right)^n=0, $$ so $$ \lim_{n\to\infty}\left(\sin\left(\frac{2\pi n}{3n+1}\right)\right)^n=0. $$

Answer 2

To do it rigorously first say that since $a_n=\sin(\frac{2n\pi}{3n+1})\to \frac{\sqrt{3}}{2}$, for all large $n$ , $0<a_n\leq c <1$.

Hence $(a_n)^n\leq c^n$ for all large $n$ and as the right hand side goes to $0$, you have the desires result that the original sequence goes to $0$

Answer 3

Let $$A_n=\left(\sin\left(\frac{2\pi n}{3n+1}\right)\right)^n \quad \implies \quad \log(A_n)=n \log\left(\sin\left(\frac{2\pi n}{3n+1}\right)\right)$$ Using Taylor series $$\sin\left(\frac{2\pi n}{3n+1}\right)=\frac{\sqrt{3}}{2}+\frac{\pi }{9 n}+O\left(\frac{1}{n^2}\right)$$ $$\log\left(\sin\left(\frac{2\pi n}{3n+1}\right)\right)=-\frac{1}{2} \log \left(\frac{4}{3}\right)+\frac{2 \pi }{9 \sqrt{3} n}+O\left(\frac{1}{n^2}\right)$$ $$\log(A_n)=-\frac{1}{2} \log \left(\frac{4}{3}\right)n+\frac{2 \pi }{9 \sqrt{3}}+O\left(\frac{1}{n}\right)$$ $$A_n=e^{\log(A_n) }=\exp\left(-\frac{1}{2} \log \left(\frac{4}{3}\right)n+\frac{2 \pi }{9 \sqrt{3}}+O\left(\frac{1}{n}\right) \right)$$