# 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.

• Are you sure that the problem is not $$x_n=\prod^{n}_{i=2}(1+\pi^{1-2\color{red}{i}})$$ Oct 29, 2021 at 10:32
• @ClaudeLeibovici Yes you are right. I got a typo in it Oct 29, 2021 at 11:44
• Here is a quite similar problem: math.stackexchange.com/q/3206623/42969. Oct 29, 2021 at 13:09

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 $$(x_\infty-1)$$ $$1+\frac{11}{100} \sqrt{\sqrt[3]{6}-\sqrt[3]{5}}=\color{red}{1.0360062}52$$