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

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.

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:

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