2006 SMT/Team Problems/Problem 15

Solution

Denote $\prod_{i=0}^{\infty} \frac{c_i^{2}}{c_i^{2}-1}$ as $\prod_{composite} \frac{c^2}{c^2-1}$ . To begin, let $S$ = $\prod_{n=2}^{\infty} \frac{n^2}{n^2-1}$ .

Let's notice that:

$\begin{align*} \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} \frac{n^2-1}{n^2}\\ \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} (1-\frac{1}{n^2})\\ \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} (1-n^{-2})\\ \end{align*}$

And that:

$\begin{align*} \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} \frac{n^2-1}{n^2}\\ \frac{1}{S} &= \lim_{k\to\infty} \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^2}\\ \frac{1}{S} &= \frac{1\cdot3}{2\cdot2} \cdot \frac{2\cdot4}{3\cdot3}\cdots \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot\frac{(k-1)\cdot(k+1)}{k\cdot k} \end{align*}$

Notice that all the fractions, except $\frac{1}{2}$ and $\frac{k+1}{k}$ will cancel out with their reciprocals from the next term(i.e $\frac{3}{2}$ cancels with $\frac{2}{3}$ , $\frac{k}{k-1}$ with $\frac{k-1}{k}$ ).

Therfore:

$\begin{align*} \frac{1}{S} &= \lim_{k\to\infty} \frac{1}{2} \cdot \frac{k+1}{k}\\ \frac{1}{S} &= \lim_{k\to\infty} \frac{1}{2} \left(1+\frac{1}{k}\right)\\ \frac{1}{S} &=\frac{1}{2} \end{align*}$

So we have now proven that $\frac{1}{S} = \frac{1}{2}$ . We will save this for later. For the second part, we will use the famous identity (discovered by Euler) that: $\prod_{primes} (1-p^{-s}) = \frac{1}{\xi(s)}$ .

Plugging in $s=2$ , we see that: $\prod_{primes} (1-p^{-2}) = \frac{1}{\xi(2)} = \frac{6}{\pi^2}$ .

Therefore, we can split $\frac{1}{S}=\frac{1}{2}$ into two infinite products, one of prime numbers, and one of composite numbers $\begin{align*} \frac{1}{S} &= \prod_{n=2}^{\infty} (1-n^{-2})\\ \frac{1}{2} &= \prod_{primes} (1-p^{-2}) \cdot \prod_{composite} (1-c^{-2})\\ \frac{1}{2} &= \frac{1}{\xi(2)}\cdot \prod_{composite} (1-c^{-2})\\ \frac{1}{2} &= \frac{6}{\pi^2}\prod_{composite} (1-c^{-2})\\ \frac{\pi^2}{12} &= \prod_{composite} (1-c^{-2}) \end{align*}$

Noticing that:

$\prod_{composite} (1-c^{-2}) = \prod_{composite} (1-\frac{1}{c^2}) = \prod_{composite} \frac{c^2-1}{c^2}$

We can conclude that: $\begin{align*} \frac{\pi^2}{12} &= \prod_{composite} \frac{c^2-1}{c^2}\\ \frac{12}{\pi^2} &= \prod_{composite} \frac{c^2}{c^2-1} \end{align*}$

Therefore, the answer is $\boxed{\frac{12}{\pi^2}}$ , $\emph{Dividend}$

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2006 SMT/Team Problems/Problem 15

Solution