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.

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

1 Answer 1


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

  • $\begingroup$ You can mention these are q-pochhammer symbols.Maybe the symbols can simplify into another q-function? $\endgroup$ Oct 29, 2021 at 13:06
  • $\begingroup$ Thanks a lot for your wonderful solution. It means a lot to me. $\endgroup$ Oct 29, 2021 at 13:06
  • $\begingroup$ @TymaGaidash. What are my very first words . $\endgroup$ Oct 29, 2021 at 13:08
  • $\begingroup$ Note the “q” in the link $\endgroup$ Oct 29, 2021 at 13:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .