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?
 A: 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
$$
\prod_{n=1}^{N-1}\frac{T_n}{T_n+1} = \frac{2\pi}{\cosh\left( \frac{\sqrt 7}{2}\pi\right)} \cdot \frac{(N-1)! N!}{\Gamma\left(N+\frac 12 + \frac{\sqrt 7}{2i}\right)\Gamma\left(N + \frac 12 - \frac{\sqrt 7}{2i}\right)} \, .
$$
A: You can also do it using Pochhammer symbols
$$\frac{n(n+1)}{n(n+1)+2}=\frac{n(n+1)}{(n-a)(n-b)}$$
$$P_p=\prod_{n=1}^p\frac{n(n+1)}{(n-a)(n-b)}=\frac{\prod_{n=1}^p n(n+1) }{{\prod_{n=1}^p (n-a)(n-b)} }=\frac{\Gamma (p+1) \Gamma (p+2)}{(1-a)_p (1-b)_p}$$
Take logarithms and use Stirling approximation
$$\log(P_p)=((a+b) \log (p)+\log (p \Gamma (1-a) \Gamma (1-b)))+\frac{-a^2+a-b^2+b+2}{2
   p}+O\left(\frac{1}{p^2}\right)$$ Now, using $a=\frac{1}{2} \left(-1-i \sqrt{7}\right)$ and $b=\frac{1}{2} \left(-1+i \sqrt{7}\right)$
$$\log(P_p)=\log \left(2 \pi  \text{sech}\left(\frac{\sqrt{7} \pi
   }{2}\right)\right)+\frac{2}{p}+O\left(\frac{1}{p^2}\right)$$
$$P_p=e^{\log(P_p)}=2 \pi  \text{sech}\left(\frac{\sqrt{7} \pi }{2}\right)+\frac{4 \pi 
   \text{sech}\left(\frac{\sqrt{7} \pi }{2}\right)}{p}+O\left(\frac{1}{p^2}\right)$$
Edit
Making the problem more general
$$Q_p=\prod_{n=1}^p\frac{n(n+1)}{n(n+1)+k}$$ and using the same procedure
$$Q_p=\pi  k \,\text{sech}\left(\frac{\pi }{2}  \sqrt{4 k-1}\right)\exp\Bigg[ \frac{k}{p}-\frac{k}{p^2}-\frac{(k-6) k}{6 p^3}+\frac{(k-2) k}{2 p^4}+O\left(\frac{1}{p^5}\right)\Bigg]$$
A: 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:
$$\frac{2\pi}{\cosh(\frac{\pi\sqrt{7}}{2})} = \prod_{n=1}^\infty\frac{n(n+1)}{n(n+1)+2} = \prod_{n=1}^\infty \frac{T_n}{T_n + 1}$$
