To calculate the infinite product $$\prod_{n=1}^{\infty}\frac{2}{1+\pi^{\frac{1}{2^n}}},$$ I found that $\sum_{n=1}^{\infty}\ln\left(\frac{2}{1+\pi^{\frac{1}{2^n}}}\right)$ converges, so as $\prod_{n=1}^\infty\frac{2}{1+\pi^{\frac{1}{2^n}}}.$   The task though is asking to calculate the product  and not just verify the convergence. Is there a mathematical way to calculate this product, or calculate this limit : $$\lim_{n\rightarrow\infty}\frac{2^n}{\left(1+\pi^{\frac{1}{2}}\right)\left(1+\pi^{\frac{1}{2^2}}\right)\cdots\left(1+\pi^{\frac{1}{2^n}}\right)}$$  Thank you for your time
I run into this infinite product today. I don't have that much experience evaluating products therefore I don't have any idea of how to tackle this.   Here is the question. Evaluate (if possible) in a closed form the product:  $$\Pi = \prod_{n=1}^{\infty} \frac{1}{1+\pi^{{\large \frac{1}{2^n}}}}$$  The numerical value seems to be $\Pi= 0.534523$ which is very close to $\frac{\pi}{6}$ taking into account that $\frac{\pi}{6} \approx 0.523598$. In the mean time W|A evaluates it to $0$. I'm lost.   Can the community help?  Edit: Based on the answer we have two products. The product I asked tends to zero and the bonus product provided by @you're in my eye (thanks for that) is $\frac{\ln \pi}{\pi-1}$.   Thanks for the quick response.
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