Prove that the sequence $\{x_n\}$ converges, where: $x_n = 2^n \sin\left(\frac{\pi}{2^{n+1}}\right)$

Question

I think it converges to $\frac{\pi}{2}$, but I am not sure how to prove it. I've tried using induction on $n$ but have had no luck. Any help would be appreciated thanks.

Answer 1

Use $\sin x=x+O(x^3)$ as $x\to0$. Then $$\sin\frac\pi{2^{n+1}}=\frac{\pi}{2^{n+1}}+O(2^{-3n})$$ and $$2^n\sin\frac\pi{2^{n+1}}=\frac{2^n\pi}{2^{n+1}}+O(2^{-2n})$$ etc.

Answer 2

use that $|\sin x| < |x|$

$\sin \frac{\pi}{2^{n+1}} < \frac{\pi}{2^{n+1}}$

$0<2^{n}\sin \frac{\pi}{2^{n+1}} < \frac {\pi}{2}$

The sequence is bounded... can we show that it is monotone?

$x_n = $$2^{n}\sin \frac{\pi}{2^{n+1}}\\ 2^{n}\sqrt {\frac {1-\cos \frac{\pi}{2^{n}}}{2}}\\ 2^{n}\frac {\sin \frac {\pi}{2^n}}{\sqrt {2(1+\cos \frac{\pi}{2^{n}})}}\\ x_{n-1}\sqrt {\frac {2}{1+\cos \frac{\pi}{2^{n}}}}\\ \sqrt {\frac {2}{1+\cos \frac{\pi}{2^{n}}}}>1$

$\frac {x_{n+1}}{x_{n}} > 1$ suggests that it is.