<s>[INST] Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:

$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \frac{1}{1+\frac 2{n(n+1)}}$$

Searching on MSE I've found solutions for other similar problems related to Weierstrass' product for the sine function:
$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)$$
but nothing specific on that particular case.
I've tried to use Weierstrass' product but without success.
Also expansion in logarithmic sum allow me to find good estimation with a small number of terms but nothing more than that.
According to wolfram the value for the infinite product should be: $\frac{2\pi}{\cosh\left(\frac{\sqrt 7}2\pi\right)} \approx 0.197$.
How can we determine the value for the partial and the infinite product in a closed form?
 [/INST] Start from the infinite product expansion of normalized sinc function:
$${\rm sinc}(z) =\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \left(1 - \frac{z^2}{n^2}\right)
$$
Substitute $z$ by $2z$, grouping factors in RHS in pairs and then divide it by expansion of ${\rm sinc}(z)$, we obtain an infinite product expansion of $\cos \pi z$:
$$\cos\pi z = \frac{{\rm sinc}(2z)}{{\rm sinc}(z)}
= \prod_{k=1}^\infty\left(1 - \frac{z^2}{(k-\frac12)^2}\right)
$$
Move first factor on RHS to LHS and let $n = k - 1$, this becomes
$$\frac{\cos\pi z}{1 - 4z^2} = \prod_{n=1}^\infty\left(1 - \frac{z^2}{(n+\frac12)^2}\right)\tag{*1}$$
Taking limit at $z = \frac12$ and apply L'Hopital's rule, LHS becomes
$$\lim_{z\to\frac12} \frac{\cos\pi z}{1 - 4z^2} = \lim_{z\to\frac12}
\frac{-\pi \sin\pi z}{-8z} = \frac{\pi}{4}$$
This leads to
$$\frac{\pi}{4} = \prod_{n=1}^\infty\left(1 - \frac{\frac14}{(n+\frac12)^2}\right) = \prod_{n=1}^\infty\frac{n(n+1)}{(n+\frac12)^2}\tag{*2a}$$
In $(*1)$, substitute $z$ by $i\frac{\sqrt{7}}{2}$, we obtain
$$\frac{\cosh(\frac{\pi\sqrt{7}}{2})}{8} = \prod_{n=1}^\infty\left(1 + \frac{\frac74}{(n+\frac12)^2}\right)
= \prod_{n=1}^\infty \frac{n(n+1)+2}{(n+\frac12)^2}\tag{*2b}$$
Divide $(*2a)$ by $(*2b)$, we obtain:
$$\frac{2\pi}{\cosh(\frac{\pi\sqrt{7}}{2})} = \prod_{n=1}^\infty\frac{n(n+1)}{n(n+1)+2} = \prod_{n=1}^\infty \frac{T_n}{T_n + 1}$$
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