Value of $\prod_{n=2}^{\infty} \frac{n^2+1}{n^2-1}$

Question

Consider the following product: $$\prod_{n=2}^{\infty} \frac{n^2+1}{n^2-1} = \prod_{n=2}^{\infty} \frac{1+\frac{1}{n}}{1-\frac{1}{n}}\approx 3.67608...$$ It seems to be close to OEIS A156648, i.e. $\prod_{n=1}^{\infty} 1+\frac{1}{n^2}$, which is also expressible as $\frac{\sinh(\pi)}{\pi}$. Is the first product expressible in similar closed form?

Answer 1

Write your product as $$\prod_{n=2}^\infty \frac{1+\frac{1}{n^2}}{1-\frac{1}{n^2}}$$ and then notice that the product of the denominators telescopes: $$\prod_{n=2}^\infty \bigg(1-\frac{1}{n^2}\bigg)=\prod_{n=2}^\infty \frac{(n+1)(n-1)}{n^2}=\frac{1}{2}$$ making your product equal to $$2\prod_{n=2}^\infty \bigg(1+\frac{1}{n^2}\bigg)=\frac{\sinh(\pi)}{\pi}$$