How Is the Infinite Product of a Recurrence Sequence Solved?

[Ackbach]

Here is this week's POTW:

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Let $a_0 = \dfrac52$ and $a_k = a_{k-1}^2 - 2$ for $k \geq 1$. Compute $\displaystyle\prod_{k=0}^\infty \left(1 - \frac{1}{a_k} \right)$ in closed form.

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[Ackbach]

No one answered this week's POTW, which was Problem A-3 in the 2014 Putnam Archives. The solution, attributed to Kiran Kedlaya and associates, follows.

[sp]
Using the identity
\[
(x + x^{-1})^2 - 2 = x^2 + x^{-2},
\]
we may check by induction on $k$ that $a_k = 2^{2^k} + 2^{-2^k}$; in particular, the product is absolutely convergent. Using the identities
\begin{align*}
\frac{x^2 + 1 + x^{-2}}{x + 1 + x^{-1}} &= x - 1 + x^{-1}, \\
\frac{x^2 - x^{-2}}{x - x^{-1}} &= x + x^{-1},
\end{align*}
we may telescope the product to obtain
\begin{align*}
\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}} \\
&= \prod_{k=0}^\infty \frac{2^{2^{k+1}} + 1 + 2^{-2^{k+1}}}{2^{2^k} + 1 + 2^{-2^k}} \cdot
\frac{2^{2^k} - 2^{-2^k}}{2^{2^{k+1}} - 2^{2^{-k-1}}} \\
&= \frac{2^{2^0} - 2^{-2^0}}{2^{2^0}+1 + 2^{-2^0}} = \frac{3}{7}.
\end{align*}
[/sp]