<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] For complex $a, b, c, d$ with non-negative real part we have
$$
 \prod_{n=1}^{N-1}\frac{(n+a)(n+b)}{(n+c)(n+d)} = \frac{\Gamma(c+1)\Gamma(d+1)}{\Gamma(a+1)\Gamma(b+1)} \cdot \frac{\Gamma(a+N)\Gamma(b+N)}{\Gamma(c+N)\Gamma(b+N)} \, .
$$
If additionally  $a+b=c+d$ then Stirling's formula shows that the second factor converges to $1$ for $N \to \infty$, so that
$$
\prod_{n=1}^{\infty}\frac{(n+a)(n+b)}{(n+c)(n+d)} = \frac{\Gamma(c+1)\Gamma(d+1)}{\Gamma(a+1)\Gamma(b+1)} \, .
$$
In our case,
$$
 \prod_{n=1}^{\infty}\frac{T_n}{T_n+1} =  \prod_{n=1}^{\infty}\frac{n(n+1)}{\left(n+\frac 12 + \frac{\sqrt 7}{2}i\right)\left(n+\frac 12 - \frac{\sqrt 7}{2}i\right)} \\
= \Gamma\left(\frac 32 + \frac{\sqrt 7}{2i}\right)\Gamma\left(\frac 32 - \frac{\sqrt 7}{2i}\right) \\
= \left(\frac 12 + \frac{\sqrt 7}{2}i\right)\left(\frac 12 - \frac{\sqrt 7}{2}i\right)\Gamma\left(\frac 12 + \frac{\sqrt 7}{2}i\right)\Gamma\left(\frac 12 - \frac{\sqrt 7}{2}i\right) \\
= 2 \frac{\pi}{\sin\left(\pi\left(\frac 12 + \frac{\sqrt 7}{2}i\right)\right)}
= \frac{2\pi}{\cosh\left( \frac{\sqrt 7}{2}\pi\right)} \, ,
$$
using the functional equation of the Gamma function and Euler's reflection formula. For the partial product we then get
$$
\prod_{n=1}^{N-1}\frac{T_n}{T_n+1} = \frac{2\pi}{\cosh\left( \frac{\sqrt 7}{2}\pi\right)} \cdot \frac{(N-1)! N!}{\Gamma\left(N+\frac 12 + \frac{\sqrt 7}{2i}\right)\Gamma\left(N + \frac 12 - \frac{\sqrt 7}{2i}\right)} \, .
$$
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