# Find $\lim _{r\to \infty}\frac{\left(\prod_{n=1}^{r}\sin\left(nx\right)\right)}{\left(\frac{1}{r}\right)}$

$$\lim _{r\to \infty}\frac{\left(\prod_{n=1}^{r}\sin\left(nx\right)\right)}{\left(\frac{1}{r}\right)}$$

I tried using the product sin formula but got nowhere.& even after multipling and dividing by $$2cos (x)$$, answer couldnt be obtained as only multiples of two cut out themselves.
it is a 0/0 indeterminate form.
Also I didn't get the answer by using the l's-hopital rule.
I even tried graphing it on desmos, but the graph was strange---( I think even desmos couldn't compute it further)

• You seem to have the wrong function, the term in the product $\prod_{n-1}^r\sin(rx)$ doesn't depend on $n$... May 8 '20 at 10:36
• thanks. corrected that May 8 '20 at 11:17
• The product is zero if $x$ is a rational multiple of $\pi$. This would lead me to believe the product tends to zero for any real $x$ by a density argument, though I can't quite prove it. I don't believe the product has a known closed-form either, which doesn't help. May 10 '20 at 1:58

If $$x=\frac pq \pi$$ for $$p,q\in \mathbb Z, (q\ne 0)$$, then the limit would evaluate to zero as for any $$q$$ we have the term $$\sin(qx)$$ in our product. If $$x$$ is not a rational multiple of $$\pi$$, then no term in the numerator equals zero and we have to take the limit.

$$\because -1 \lt \sin(mx) \lt 1$$ doesn’t vary much, the limit can be reasonably approximated as:

$$\lim_{r\to\infty} r\prod_{n=1}^r \sin(nx) \sim \lim_{r\to\infty} r (\sin x)^r \\ =\lim_{r\to\infty} \frac{r}{(\csc x)^r} =0$$

$$\because|\csc x| \gt 1$$, the limit equals zero due to the fact exponentials grow way faster.

• Having spent some time on this, could you please elaborate on passing the product to $r$ powers of $-\sin(x)$? May 10 '20 at 22:06
• @Integrand Sorry, the $(-1)^r$ wasn’t meant to be there. May 10 '20 at 22:38
• Even the passing to $r$ powers is not obvious. For instance, the locations of extrema and zeroes change. May 10 '20 at 23:37
• @Integrand That’s true, but it’s a good approximation , especially as $r$ gets large. The idea is that a bunch of numbers ranging from $-1$ to $1$ when multiplied together infinitely many times produce the same result as multiplying one particular number among them, infinitely many times. May 11 '20 at 9:28

If $$x$$ is a rational multiple of $$\pi$$, then for some integer $$N > 0$$, $$\sin(Nx) = 0$$. This forces $$\prod_{n=1}^r \sin(nx) = 0$$ whenever $$r \ge N$$. In this case, the limit is $$0$$.

Otherwise, $$x$$ is not a rational multiple of $$\pi$$ and $$|\cos x| < 1$$. Notice

$$|\sin(nx)\sin(n+1)x| = \frac{|\cos x - \cos((2n+1)x)|}{2} \le \mu \stackrel{def}{=}\frac{1 + |\cos x|}{2}$$

By grouping the factors in the numerator in pairs, we have following bound for the weighted product at finite $$r$$.

$$r\left|\prod_{n=1}^r \sin(nx)\right| \le r\prod_{k=1}^{\lfloor r/2\rfloor} |\sin((2k-1)x)\sin(2kx)| \le r\mu^{\lfloor r/2\rfloor}$$ Since $$\mu < 1$$, the limit of the weighted product is again $$0$$.