$\prod_{n=1}^{\infty} n \sin(1/n)$

Since:

$A=\prod_{n=1}^{\infty} n \sin(1/n)=\prod_{n=1}^{\infty} \prod_{m=1}^{\infty} \left(1-\frac{1}{n^{2}m^{2}\pi^{2}}\right)$

$\log(A) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \log\left(1-\frac{1}{n^{2}m^{2}\pi^{2}}\right)$

$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}}{v} \left[-\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}$

$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v} \left[\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}$

$\sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v}\frac{1}{\pi^{2v}} \left(\sum_{n=1}^{\infty} \frac{1}{n^{2v}}\right) \left(\sum_{m=1}^{\infty} \frac{1}{m^{2v}}\right)$

$\sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v}\frac{\zeta(2v)^{2}}{\pi^{2v}}$

(which can easily be evaluated in wolfram alpha, and converges much faster than the original double product)

Therefore:

$\prod_{n=1}^{\infty} n \sin(1/n)\approx 0.755363388518573214063...$