# Limit of product of $\sin \frac{k}{n}$

Could you help me how to find the limit of $$\left(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1\right)^{\frac{1}{n}}?$$

I know that $$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$

and $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right) = \int_0^1 \ln(\sin(x)) \, dx \text{ (Riemann integral)}$$

but I am not sure what to do next, I mean, how do I get back to $$\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1)^{\frac{1}{n}}?$$

Could you help me with that?

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Just exponentiate the result of the integration. – Ron Gordon Jun 4 '13 at 17:49
And be carefull. $\int_0^1 \ln(\sin(x)) dx$ is not a Riemann Intregral. It is an improper integral. See discussions here: math.stackexchange.com/questions/406819/… – N. S. Jun 4 '13 at 17:50
Is it $e^{\int_0 ^1 \ln (\sin x) dx}$? – Hagrid Jun 4 '13 at 18:08
@Hagrid: Yes, it is. – Mhenni Benghorbal Jun 4 '13 at 19:04
@Hagrid: If you are interested the integral, you can use the technique. – Mhenni Benghorbal Jun 4 '13 at 19:27

I am not quite sure, but maybe you do not need integration: let us consider $$exp \left( ln \left(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1\right)\frac{1}{n} \right).$$ Now let us expand the sinus: $$exp \left( ln \left( \frac{(n-1)!}{n^{n-1}}+ o\left( \frac{1}{n^{n-1}}\right) \right)\frac{1}{n} \right)=exp \left(\frac{n-1}{n}ln\left(\frac{n-1}{en}\right)+o(1) \right) \rightarrow \frac{1}{e}.$$ Here I used the Stirling approximation. I hope I haven't done anything wrong!

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