Convergence and value of infinite product $\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)$?

Since the limit $\frac{\sin(x)}{x}=1$ for $x \rightarrow 0$, I wondered about the infinite product:

$$\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)=\sin(1) \cdot 2 \sin\left( \frac{1}{2} \right) \cdot 3 \sin\left( \frac{1}{3} \right) \dots$$

By numerical experiment in Mathematica it seems to converge, even if very slowly (I mean to non-zero value):

$$P(14997)= 0.755371783$$

$$P(14998)= 0.755371782$$

$$P(14999)= 0.755371782$$

$$P(15000)= 0.755371781$$

I can prove the convergence by integral test for the series:

$$\sum^{\infty}_{n=1} \ln\left( n \sin \left( \frac{1}{n} \right) \right)$$

$$\int^{\infty}_{1} \ln\left( x \sin \left( \frac{1}{x} \right) \right) dx=\int^{1}_{0} \frac{1}{y^2} \ln \left( \frac{\sin (y)}{y} \right) dy=-0.168593$$

I think the integral test can work with negative function as long as it's monotone, otherwise I can just put the minus sign before the infinite sum.

By the way, this is a related question about the convergence of the sum above.

But I'm more interested in the infinite product itself.

I'm not sure if the value of this infinite product can be found and how to go about it. Is it zero or not? Any thoughts would be appreciated

Solution 1:

Thanks for all the comments. The closed form is unlikely, but some very good estimates can be provided, using Taylor series for $\sin$:

$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<\prod_{n=1}^{\infty} 1=1$$

$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}>\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2} \right)=\frac{\sqrt{6}}{\pi} \sin \frac{\pi}{\sqrt{6}}=0.747529$$

$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2}+\frac{1}{120n^4} \right)< \exp \left(-\frac{\pi^2}{36}+\frac{\pi^4}{10800} \right)=0.767101$$

Actually, for the last product Mathematica gives the closed form:

$$\prod_{n=1}^{\infty} \left(1-\frac{1}{6n^2}+\frac{1}{120n^4} \right)=\frac{5\sqrt{5}}{\pi^2} \sin \left(\frac{\pi \sqrt{1-\frac{i}{\sqrt{5}}}}{2 \sqrt{3}}\right) \sin \left(\frac{\pi \sqrt{1+\frac{i}{\sqrt{5}}}}{2 \sqrt{3}}\right)=$$

See this answer.

$$=\frac{\sqrt{30}}{\pi^2} \left(\cosh \left( \pi \sqrt{\frac{1}{\sqrt{30}}-\frac{1}{6}} \right)-\cos \left( \pi \sqrt{\frac{1}{\sqrt{30}}+\frac{1}{6}} \right) \right)=0.755542$$

We get the estimation:

$$\prod_{n=1}^{\infty} n \sin \frac{1}{n}<0.755542$$

Which gives at least three (maybe four) correct digits for the numerical value.

Solution 2:

Alright, this is another answer, much better one.

We can evaluate this product numerically with excellent precision, if we get it into a better form.

$$P=\prod^{\infty}_{n=1} n \sin \left( \frac{1}{n} \right)=\prod^{\infty}_{n=1} \prod^{\infty}_{k=1} \left(1- \frac{1}{\pi^2 n^2 k^2} \right)$$

Now we take logarithm of the product:

$$\ln P=\sum^{\infty}_{n=1} \sum^{\infty}_{k=1} \ln \left(1- \frac{1}{\pi^2 n^2 k^2} \right)=-\sum^{\infty}_{n=1} \sum^{\infty}_{k=1}\sum^{\infty}_{l=1}\frac{1}{l~\pi^{2l} n^{2l} k^{2l}}=-\sum^{\infty}_{l=1}\frac{\zeta (2l)^2}{l~\pi^{2l}}$$

This last single sum Mathematica computes with great precision, so we can write:

$$\ln P=-0.280556336229155079602039680939198362173$$

And the product is:

$$P=\exp \left(-\sum^{\infty}_{l=1}\frac{\zeta (2l)^2}{l~\pi^{2l}} \right)=0.75536338851857321406336498617047655360$$

By the same logic we also have:

$$P_1=\prod^{\infty}_{n=1} n \sinh \left( \frac{1}{n} \right)=\exp \left(-\sum^{\infty}_{l=1}\frac{(-1)^l \zeta (2l)^2}{l~\pi^{2l}} \right)=1.307970936664283649012104476$$