# 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

Here's a way to show that the right Riemann sum $\displaystyle \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$ converges to the value of the improper integral $\displaystyle \int_{0}^{1} \ln (\sin x) \ dx$ :

The function $-\ln (\sin x)$ is nonnegative and decreasing on $(0,1]$.

Let $\displaystyle S_{n} = \frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$.

Then $- \displaystyle S_{n} < -\int_{0}^{1} \ln (\sin x) \ dx < - \int_{0}^{\frac{1}{n}} \ln (\sin x) \ dx - S_{n}$

$\displaystyle \implies 0 < -\int_{0}^{1} \ln (\sin x) \ dx + S_{n} < -\int_{0}^{\frac{1}{n}} \ln(\sin x) \ dx$

$\displaystyle \implies 0 < -\int_{0}^{1} \ln (\sin x) \ dx + \lim_{n \to \infty} S_{n} < 0$

$\displaystyle \implies 0 < \int_{0}^{1} \ln (\sin x) \ dx - \lim_{n \to \infty} S_{n} < 0$

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Could you explain to me why $0 < -\int_{0}^{1} \ln (\sin x) \ dx + S_{n} < -\int_{0}^{\frac{1}{n}} \ln(\sin x) \ dx \implies 0 < -\int_{0}^{1} \ln (\sin x) \ dx + \lim_{n \to \infty} S_{n} < 0$? –  Hagrid Jun 5 '13 at 5:37
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!