<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] 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
$$Q_p=\pi  k \,\text{sech}\left(\frac{\pi }{2}  \sqrt{4 k-1}\right)\exp\Bigg[ \frac{k}{p}-\frac{k}{p^2}-\frac{(k-6) k}{6 p^3}+\frac{(k-2) k}{2 p^4}+O\left(\frac{1}{p^5}\right)\Bigg]$$
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