Putnam 2014, Problem A-3

Let $a_0 = 5/2$ and $a_k = a_{k-1}^2 - 2$ for $k \geq 1$. Compute
$$\prod_{k=0}^{\infty} \left(1 - \frac{1}{a_k}\right)$$
in closed form.

Solution:
First solution: Using the identity \((x + x^{-1})^2 - 2 = x^2 + x^{-2}\), we check by induction that \(a_k = 2^{2^k} + 2^{-2^k}\). The product is absolutely convergent. Using the identities  
\[
\frac{x^2 + 1 + x^{-2}}{x + 1 + x^{-1}} = x - 1 + x^{-1}, \quad \frac{x^2 - x^{-2}}{x - x^{-1}} = x + x^{-1},
\]  
we telescope the product to obtain  
\[
\prod_{k=0}^\infty \left( 1 - \frac{1}{a_k} \right) = \prod_{k=0}^\infty \frac{2^{2^k} - 1 + 2^{-2^k}}{2^{2^k} + 2^{-2^k}} = \frac{2^{2^0} - 2^{-2^0}}{2^{2^0} + 1 + 2^{-2^0}} = \frac{3}{7}.
\]  
Second solution: (by Catalin Zara) The sequence \(a_k\) is increasing and unbounded. Using \(a_{k+1} + 1 = (a_k - 1)(a_k + 1)\), we check by induction that  
\[
\prod_{k=0}^n \left( 1 - \frac{1}{a_k} \right) = \frac{2}{7} \cdot \frac{a_{n+1} + 1}{a_0 a_1 \cdots a_n}.
\]  
Using \(a_{n+2}^2 - 4 = a_{n+1}^4 - 4a_{n+1}^2\), we get \(a_0 a_1 \cdots a_n = \frac{2}{3} \sqrt{a_{n+1}^2 - 4}\). Hence  
\[
\prod_{k=0}^n \left( 1 - \frac{1}{a_k} \right) = \frac{3}{7} \cdot \frac{a_{n+1} + 1}{\sqrt{a_{n+1}^2 - 4}} \to \frac{3}{7} \quad \text{as } n \to \infty.
\]