# Evaluate an infinite product in a closed form

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.

• The product seems to be zero. How did you get a non-zero value? – Yuriy S Jul 27 '16 at 20:33
• Well from one perspective one of Wolfram Alpha online calculators says that $\Pi$ is equal to that value. Wolfram Alpha main page says it's zero. But zero does not stand well to me. Anyway, I don't have a solution neither do I know the final answer. So, I am waiting for the community to share its thoughts. – Tolaso Jul 27 '16 at 20:36
• There must be some discrepancy between what you're calculating and what your statement depicts. Each of the fractions in this infinite product is less than $\frac{1}{2}$, so surely the limiting product must be $0$. Am I missing something? ¶ ETA: Ahh, a missing factor of $2$ makes a big difference. – Brian Tung Jul 27 '16 at 20:38

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.)

• Yes, OK.. !! no problem .. Updated my answer. (+1) anyway. – Tolaso Jul 27 '16 at 20:50

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

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.

• (+1) for the formula. I would really like a derivation of it. – Tolaso Jul 27 '16 at 20:39
• @user1952009, I did in the first comment. It's obvious the OP was computing the product I provided – Yuriy S Jul 27 '16 at 20:42
• why don't you mention his product diverges to $0$ in your answer ? – reuns Jul 27 '16 at 20:44
• OK, now we have an answer to both product. Thank you all. – Tolaso Jul 27 '16 at 20:46
• @You'reInMyEye I'm accepting the other answer. Sorry for my confusion. Anyway thanks for your time. – Tolaso Jul 27 '16 at 20:53

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)

• Actually we cannot use DLH to find the limit of the last sequence. What we can do instead is to use the already formula: $$\frac{1}{\ln x}= \lim_{n \rightarrow +\infty} \frac{1/n}{1-x^{1/n}}$$ and combine them all together to the that the final answer is $\frac{\ln \pi}{\pi-1}$. – Tolaso Jul 27 '16 at 21:59