<s>[INST] Evaluating $\lim_{n\to \infty}(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n})^{1/n}$ Find:
$$\lim_{n\to \infty}\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n}$$
I was wondering,is there a formula for multiplication of sines,when angles are in arithmetic progression(likewise there is formula for summation of sines).I also tried this question by taking log of both sides and then applying L Hospital rule.But it is becoming messy.Can someone please assist me in solving this question? 
 [/INST] We have
$$\lim_{n\to\infty}\ln \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =\lim_{n\to\infty}\sum_{k=1}^{n-1} \frac{1}{n}\ln \sin\left(\frac{k}{n}\cdot \frac{\pi}{2} \right) =\hspace{1cm}\int_{0}^{1} \ln \sin\frac{\pi t}{2} dt   $$
Hence $$\lim_{n\to\infty} \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =e^{\int_0^1\ln \sin\frac{\pi t}{2} dt}  $$
But it can be calculated that $$\int_0^1\ln \sin\frac{\pi t}{2} dt =-\ln2$$ therefore $$\lim_{n\to\infty} \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =e^{-\ln 2 }=\frac{1}{2}  $$
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