# Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$

Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?

• are the values in the sines in degree or radiant?
– CODE
May 30, 2013 at 15:38
• @ThomasAndrews I made a mistake. May 30, 2013 at 15:41
• Yes, unless you have strong reason to believe these are related, you might want to separate it into two problems. May 30, 2013 at 15:46
• Just a suggestion: you may want to compute out the values for a large number of values for n, plot the results in various ways, and see if maybe a limiting value appears to be achieved, from which may give you a hint as to what the true answer is. Dec 14, 2013 at 12:36

few thoughts on the first one:

$$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$

This is just a Riemann sum, and thus its limit is $$\int_{0}^1 \ln(\sin(x)) dx$$

This is an improper integral though, so the RS approach might not be best, but I think it is a convergent improper integral, since $\int_0^1 \ln(x)dx$ is convergent..

Maybe someone can take over....

• I seems no closed form for $\int_0^1$ cf. math.stackexchange.com/questions/354795/… May 30, 2013 at 15:49
• It works. The improper integral is indeed convergent. And since the integrand $f(x)=\ln\sin x$ is increasing: $S_n-f(1)/n=\frac{1}{n}\sum_{k=1}^{n-1}f(k/n)\leq \int_{1/n}^1f(x)dx\leq \frac{1}{n}\sum_{k=2}^{n}f(k/n)=S_n-f(1/n)/n$. It might be better to take $-f$ which is positive decreasing, but that's the idea. May 30, 2013 at 15:58
• Another way to do it is to claim that $f(x)=\ln \sin x$ is integrable on $(0,1)$ and to see these sums as the integrals of step functions which converge pointwise to $f$. Then the result follows by dominated convergence. Or simply monotone convergence in this case. May 30, 2013 at 16:06
• But note: for an improper integral, even a convergent one, the Riemann sums may not converge to the integral. (There was a recent Monthly problem like that.) So some care is needed. May 30, 2013 at 17:09

Yes, some care is needed in order to prove the convergence of the Riemann sum to the corresponding integral, from which $$\lim_{n\to +\infty}\left(\prod_{k=1}^{n}\sin\frac{k}{n}\right)^{1/n}=\exp\left(\int_{0}^{1}\log\sin x dx\right)$$ immediately follows. The integral is just a bit less than $-1$, and we can use the Weierstrass product of the sine function to write the integral as a fast-converging series. Since: $$\sin x = x\prod_{k=1}^{+\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)$$ holds uniformly for $x\in[0,1]$, we have: $$\int_{0}^{1}\log\sin x\,dx = -1+\sum_{k=1}^{+\infty}\left(-2-2k\pi\operatorname{arctanh}\frac{1}{k\pi}+\log\left(1-\frac{1}{\pi^2 k^2}\right)\right),$$ or, using the Taylor series of the logarithm and the hyperbolic arctangent, $$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{k=1}^{+\infty}\sum_{j=1}^{+\infty}\frac{1}{j(2j+1)(k\pi)^{2j}},$$ $$\int_{0}^{1}\log\sin x\,dx = -1-\sum_{j=1}^{+\infty}\frac{\zeta(2j)}{j(2j+1)\pi^{2j}} = -\left(1+\frac{1}{18}+\frac{1}{900}+\frac{1}{19845}+\ldots\right).$$

As $x\rightarrow 0$ , $\sin(x) \rightarrow x$

So $\sin(\frac{1}{n}) \rightarrow \frac{1}{n}$ as $n \rightarrow \infty$

The product $[ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] \rightarrow \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot...\frac{n}{n} = \frac{n!}{n^n}$

As $n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n$

So $\frac{n!}{n^n} = \sqrt{2\pi n} \cdot (\frac{1}{e})^n$

$\lim_{ n \rightarrow \infty}$ of $[ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] ^ \frac{1}{n} \rightarrow \frac{1}{e}$

Since $\sin(\frac{m}{n})$ is always positive if $0< m \le n$ and less than $\frac{m}{n}$ . The limit will be less than (1/e)

• Sorry but $\sin(n/n)$ is NOT equivalent to $n/n$ and the limit is not $1/e$. Upvoter: why the upvote?
– Did
Dec 14, 2013 at 12:31