Solving infinite sums with primes.

Question

Let $p_n$ denote the $n$'th prime number.

How would one go about proving that infinite products like: $$\prod_{k=1}^\infty1 - \frac{1}{(p_k)^2} = \frac{6}{\pi^2}$$

or

$$\prod_{k=1}^\infty\frac{{p_k}^2}{{p_k}^2 - 1} = \frac{\pi^2}{6}$$ are correct?

Is there any way to prove it except by exhaustion?

Answer 1

Well,

$$\prod_{k=1}^\infty \frac{1}{1-1/p_k^2} = \prod_{k=1}^\infty \sum_{j=0}^\infty \frac1{p_k^j}.$$ Formally expanding the product of sums, and using the fundamental theorem of arithmetic, you get $$\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6.$$

But maybe that is an argument by exhaustion?

Answer 2

As @doppz mentioned in the comments, the way to prove these identities is to realize the requested product is requesting the value of $\zeta(2)$. Then, you can use any of the multitude of proofs that $\zeta(2) = \frac{\pi^2}{6}$.

For example proofs: http://math.cmu.edu/~bwsulliv/MathGradTalkZeta2.pdf

Answer 3

In general we have the following Euler formula

$$\zeta(s) = \sum_{n=1}^\infty\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}$$

and since $\zeta(2)=\frac{\pi^2}{6}$ , see this wiki

so we get the result.