Compute the series $ x_n=\prod^{n}_{i=2}(1+\pi^{1-2i}) $ The series is as following.
$$
x_n=\prod^{n}_{i=2}(1+\pi^{1-2i})
$$
The question is required to find the limit of the sequence as n tends to infinity. I have struggled for a few days and really have no idea what kind of pattern after I expand quite a number of terms.
Thank you all for any attempt to my question. I wish you all the best.

Using Pochhammer symbols $$x_n=\prod^{n}_{i=2}(1+\pi^{1-2i})=\frac{\pi }{(1+\pi )^2}\,\left(-\pi ;\frac{1}{\pi ^2}\right){}_{n+1}$$
$$x_\infty=\frac{\pi }{(1+\pi )^2}\,\left(-\pi ;\frac{1}{\pi ^2}\right){}_{\infty }$$ which is ... a number (!)
$$1.0360062490491965827813215494943798464774637545349\cdots$$ which is not recognized by inverse symbolic calculators. But, amazing or not, the $ISC$ propose, as close to it,
$$\frac{4+\log (\pi )}{4+\cos \left(\frac{\pi }{12}\right)}=\frac{4 (4+\log (\pi ))}{16+\sqrt{2}+\sqrt{6}}=\color{red}{1.036006}188$$
Still using the $ISC$ for the decimal value of $(x_\infty-1)$
$$1+\frac{11}{100} \sqrt{\sqrt[3]{6}-\sqrt[3]{5}}=\color{red}{1.0360062}52$$
