Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$

Question

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?

Answer 1

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:

$$\frac{2\pi}{\cosh(\frac{\pi\sqrt{7}}{2})} = \prod_{n=1}^\infty\frac{n(n+1)}{n(n+1)+2} = \prod_{n=1}^\infty \frac{T_n}{T_n + 1}$$

Answer 2

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 $$ \prod_{n=1}^{N-1}\frac{T_n}{T_n+1} = \frac{2\pi}{\cosh\left( \frac{\sqrt 7}{2}\pi\right)} \cdot \frac{(N-1)! N!}{\Gamma\left(N+\frac 12 + \frac{\sqrt 7}{2i}\right)\Gamma\left(N + \frac 12 - \frac{\sqrt 7}{2i}\right)} \, . $$

Answer 3

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