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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.

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  • $\begingroup$ The idea of using induction is meaningless since we are dealing with a limit of a sequence. $\endgroup$ – gimusi Nov 9 '18 at 19:39
  • $\begingroup$ I’ve voted to close that question since you are not giving sufficient context about that and it seems you have not any idea about that topic. Please add more detail if you need some advice on that. $\endgroup$ – gimusi Nov 9 '18 at 19:47
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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.

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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.

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  • $\begingroup$ Indeed in that way we are just proving the well known standard limit $\sin x/x\to 1$. $\endgroup$ – gimusi Nov 9 '18 at 19:42

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