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Question: For $n ≥ 1$, let $G_n$ be the geometric mean of $\{\sin(\frac{\pi}{2}\frac{k}{n}): 1\le k \le n\} $.

Then find $\lim_{n\to\infty}G_n $

My Approach: $G_n = \{\prod_{k=1}^n \sin(\frac{\pi}{2}\frac{k}{n})\}^{1/n}$

$\Rightarrow \ln G_n = \frac{1}{n}\{\sum_{k=1}^n \ln(\sin(\frac{\pi}{2}\frac{k}{n}))\}$

$\Rightarrow \lim_{n\to\infty} \ln G_n = \lim_{n\to\infty}\frac{1}{n}\{\sum_{k=1}^n \ln(\sin(\frac{\pi}{2}\frac{k}{n}))\}$

$\Rightarrow \lim_{n\to\infty} \ln G_n = \int_{0}^{1} \ln (\sin(\frac{\pi x}{2}))dx$

$\Rightarrow \lim_{n\to\infty} \ln G_n =\frac{2}{\pi} \int_{0}^{\pi/2} \ln (\sin(z))dz$

$\Rightarrow \lim_{n\to\infty} \ln G_n =\frac{2}{\pi}(\frac{-\pi}{2}\ln2)$

$\Rightarrow \lim_{n\to\infty} \ln G_n = -\ln 2$

$\Rightarrow G_n=\frac{1}{2}$

My problem: This has to be solved within 2-3 minutes in a competitive setting, which I don't think I'll be able to. On top of that, I have to 'remember' the specific $\int_0^{\pi/2} \ln \sin x dx=\frac{-\pi}{2}\ln 2$.

Is there a shorter process?

Tamojit Maiti
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1 Answers1

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EDIT:

By pairing up the kth term and the last kth term in $$\prod^{2^n}_{k=1}\sin(\frac{\pi k}{2^{n+1}})$$, one obtains $$=\prod^{2^{n-1}}_{k=1}\sin(\frac{\pi k}{2^{n+1}}) \sin(\frac{\pi (2^n-k)}{2^{n+1}})=\prod^{2^{n-1}}_{k=1}\sin(\frac{\pi k}{2^{n+1}}) \sin(\frac{\pi}2- \frac{\pi k}{2^{n+1}})=\prod^{2^{n-1}}_{k=1}\sin(\frac{\pi k}{2^{n+1}}) \cos(\frac{\pi k}{2^{n+1}})= \prod^{2^{n-1}}_{k=1}\frac{\sin(\frac{\pi k}{2^{n}})}2$$

END EDIT

Repeating the process for $n$ times,

$$\prod^{2^n}_{k=1}\sin(\frac{\pi k}{2^{n+1}})=\prod^{2^{n-1}}_{k=1}\frac{\sin(\frac{\pi k}{2^n})}2$$ $$...=\prod^{2^{n-n}}_{k=1}\frac{\sin(\frac{\pi k}{2^{n-n+1}})}{2^n}=\frac{\sin(\frac\pi2)}{2^n}$$

Then, $$\left(\frac{\sin(\frac\pi2)}{2^n}\right)^{1/n}=\frac12$$

Szeto
  • 11,159