Infinite product

[Tolaso J Kos]

Evaluate the product:

$$\Pi=\prod_{n=0}^{\infty} \left( 1 +\frac{1}{2^{2^n}}\right)$$

Answer

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$\Pi=2$.

[galactus]

I was kind of waiting for someone to post, but I will go ahead.

Simply note that the geometric series $$1+x+x^{2}+x^{3}+x^{4}+....=\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$$

is the same as the factorization:

$$(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\cdot\cdot\cdot (1+x^{2^{n}})=\prod_{n=0}^{\infty}(1+x^{2^{n}})=1+x+x^{2}+x^{3}+x^{4}+......$$

Since then $$\prod_{n=0}^{\infty}(1+x^{2^{n}})=\frac{1}{1-x}$$

Let $x=1/2$, and we have $$\frac{1}{1-1/2}=2$$

Come to think of it, maybe we can tie it to $$\int_{0}^{\pi}\frac{x}{x^{2}+\log^{2}(2\sin(x))}dx=2$$ to make it interesting.

[Tolaso J Kos]

Come to think of it, maybe we can tie it to $$\int_{0}^{\pi}\frac{x}{x^{2}+\log^{2}(2\sin(x))}dx=2$$ to make it interesting.

Hmm... C,

do you have a relation of this infinite product with the integral you mention? If so, that would be interesting.

P.S: A relevant integral done by RD can be found here.

[galactus]

Hi T

No, I do not have a relation between the two. I just noticed they were the same answer and thought maybe it would be cool to show.

Yes, I saw RD's solution to the other similar one. Clever.