Berkeley Math Circle

Solution:
Write the recursion as
$$\frac{a_{n+1}}{a_{n}} = \left(\frac{a_{n}}{a_{n-1}}\right)^{2} - 2$$
Let $b_{n} = a_{n+1} / a_{n}$; then $b_{n} = b_{n-1}^{2} - 2$ and $b_{0} = a$. Let
$$t = \frac{a + \sqrt{a^{2} - 4}}{2}$$
Then $a = t + 1 / t$ and it is a simple induction to show that $b_{n} = t^{2^{n}} + t^{-2^{n}}$ for $n \geq 0$. Then
$$a_{n} = a_{0} b_{0} b_{1} \cdots b_{n-1} = \prod_{i=0}^{n-1} \left(t^{2^{i}} + t^{-2^{i}}\right) = \prod_{i=0}^{n-1} \frac{t^{2^{i+1}} - t^{-2^{i+1}}}{t^{2^{i}} - t^{-2^{i}}} = \frac{t^{2^{n}} - t^{-2^{n}}}{t - t^{-1}}$$
So
$$\begin{aligned} \sum_{i=0}^{k} \frac{1}{a_{i}} & = \left(t - t^{-1}\right) \sum_{i=0}^{k} \frac{1}{t^{2^{i}} - t^{-2^{i}}} \\ & = \left(t - t^{-1}\right) \sum_{i=0}^{k} \frac{t^{2^{i}} + 1 - 1}{t^{2^{i+1}} - 1} \\ & = \left(t - t^{-1}\right) \sum_{i=0}^{k} \left(\frac{1}{t^{i} - 1} - \frac{1}{t^{2^{i+1}} - 1}\right) \\ & = \left(t - t^{-1}\right)\left(\frac{1}{t - 1} - \frac{1}{t^{2^{k+1}} - 1}\right) \\ & < \left(t - t^{-1}\right)\left(\frac{1}{t - 1}\right) \\ & = 1 + \frac{1}{t} \\ & = \frac{a + 2 - \sqrt{a^{2} - 4}}{2} . \end{aligned}$$