Evaluate an infinite product in a closed form $\prod\limits_{n=1}^{\infty} \frac{1}{1+\pi^{1/{2^n}}}$

All the terms in the product are less than half. Thus the infinite product is zero.

(The answer by You're In My Eye is for the product

$$\Pi = \prod_{n=1}^{\infty} \frac{2}{1+\pi^{{\large \frac{1}{2^n}}}}$$

The OP forgot a $2$ in the numerator of each term of the product, which is the likely cause of his confusion.)

There is a general formula:

$$\prod_{k=0}^\infty \frac{2}{1+x^{1/2^k}}=\frac{2}{x^2-1} \ln x$$

$$\prod_{k=1}^\infty \frac{2}{1+x^{1/2^k}}=\frac{1}{x-1} \ln x$$

$$\prod_{k=1}^\infty \frac{2}{1+\pi^{1/2^k}}=\frac{1}{\pi-1} \ln \pi=0.5345226992306749851 \dots$$

For the derivation of the above formula see the bottom of this answer

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Edit

The original product of the OP diverges to $0$. As I said in the first comment to the OP. But the numerical value the OP provided corresponds to this product.

The reason for divergence is in @smcc answer. I see no point in reproducing it here.

I saw your post at "Mathimatiko ergasthri".
Well, we can see that "$\frac{1}{(1+\pi^{1/2})(1+\pi^{1/4})(1+\pi^{1/8})}=\frac{(1-\pi^{1/2})(1-\pi^{1/4})(1-\pi^{1/8})}{(1+\pi^{1/2})(1+\pi^{1/4})(1+\pi^{1/8})(1-\pi^{1/2})(1-\pi^{1/4})(1-\pi^{1/8})}=\frac{(1-\pi^{1/2})(1-\pi^{1/4})(1-\pi^{1/8})}{(1-\pi^1)(1-\pi^{1/2})(1-\pi^{1/4})}=\frac{(1-\pi^{1/8})}{(1-\pi^1)}$
We can see that the $n$-th partial product is equal to $$\frac{1-\pi^{1/2^n}}{1-\pi}$$
which tends to $0$.
Now if you add a $2$ in every numerator (As I understood this was your intention from the beginning) then you want to evaluate the limit of
$$\frac{1-\pi^{1/2^n}}{1-\pi}\cdot 2^n$$ which is equal to $\frac{\ln \pi}{\pi-1}$ (Use L'Hospital)