Set $b_n=a_n/a_{n-1}$ for $n\ge 1$. Then $b_1=a$, $b_{n+1}=b_n^2-2$ and $a_n = b_1\cdots b_n$. Next, note that there is a $t>1$ such that $t+1/t =a>2$ (convince yourself of this step!). So, easy induction gives $b_n = t^{2^{n-1}} + t^{-2^{n-1}}$. Now, \[ a_n = b_1\cdots b_n = \left(t-\frac1t\right)^{-1} \left(t-\frac1t\right)\left(t+\frac1t\right)\left(t^2+t^{-2}\right)\cdots \left(t^{2^{n-1}}+t^{-2^{n-1}}\right) = \frac{t^{2^n}-t^{-2^n}}{t-t^{-1}}. \]From here, get \[ \sum_{i\ge 1}\frac{1}{a_i} = \frac1t = \frac{a-\sqrt{a^2-4}}{2}, \]as $t = \frac{a+\sqrt{a^2-4}}{2}$. So, the partial sums for any finite $k$ remain below this value, the end.

@above wow, very clever! Here is my solution, with the same recurrence but a more rocky use of it. Clearly $\frac{a_{n+1}}{a_n} = \left(\frac{a_n}{a_{n-1}}\right)^2 - 2$ and so it makes sense to set $b_n = \frac{a_{n+1}}{a_n}$ for $n\geq 0$ - clearly $b_0 = a > 2$ and $b_n = b_{n-1}^2-2$. Now for $b_n = 2c_n$ we get $c_n = 2c_{n-1}^2 - 1$. Since $c_0 = \frac{a}{2} > 1$, we may set $c_0 = \cosh \theta$ for some $\theta > 0$ and now the identity $\cosh 2x = 2\cosh^2 x - 1$ implies (by induction) $c_n = \cosh (2^n\theta)$ for all $n\geq 0$. Since $a_i = a_0b_0b_1b_2\cdots b_{i-1}$ and $a_0 = 1$, the desired inequality is reduced to $$ \sum_{i=1}^k \frac{1}{2^i\cosh \theta \cosh 2\theta \cdots \cosh 2^{i-1}\theta} < \frac{a-\sqrt{a^2-4}}{2}. $$By multiplying each numerator and each denominator by $\sinh\theta$ and use $\sinh 2x = 2\sinh x\cosh x$, we would like $$ \sum_{i=1}^k \frac{\sinh \theta}{\sinh 2^i\theta} < \frac{a-\sqrt{a^2-4}}{2}. $$Since $\theta>0$, we have $\sinh\theta >0$ and from $\cosh^2 x - \sinh^2x = 1$ we get $\sinh\theta = \frac{\sqrt{a^2-4}}{2}$, so we want $$ \sum_{i=1}^k \frac{1}{\sinh 2^i\theta} < \frac{a}{\sqrt{a^2-4}} - 1 = \coth \theta - 1. $$Since $\coth x - \coth 2x = \frac{1}{\sinh 2x}$, we get $\coth \theta = \sum_{i=1}^k \frac{1}{\sinh 2^i\theta} + \coth 2^{i+1}\theta$ and so it remains to justify $\coth 2^{i+1}\theta > 1$ (for $\theta>0$). But the definition $\coth x = \frac{e^x+e^{-x}}{e^{x}-e^{-x}}$ and $e^x > e^{-x}$ for $x>0$ clearly gives $\coth x > 1$, as desired.

Remark

Related to ISL 2003 N7 (here: https://artofproblemsolving.com/community/c6h15464p109517).

In fact we don't need to find the formula of $a_n$. Induction $a_{n+1}\geq 2a_n$ and the inequality is equivalent $4a(a^2-4)+16>0$.

@above nope, this does not work, I am quite certain! Indeed, induction shows $a_{n+1} \geq 2a_n$ for all $n$, as $\frac{a_1}{a_0} = a > 2$ and if $a_n \geq 2a_{n-1}$, then $\frac{a_n}{a_{n-1}} \geq 2$ and $\displaystyle a_{n+1} = \left(\left(\frac{a_n}{a_{n-1}}\right)^2 -2\right)a_n \geq 2a_n$. The desired inequality is equivalent to $\sum_{i=1}^k \frac{1}{a_i} < \frac{a-\sqrt{a^2-4}}{2}$. By above, the left-hand side is at most $\frac{1}{a}\sum_{i=1}^{\infty} \frac{1}{2^i} = \frac{2}{a}$ but now this is not less than the desired quantity. One can even observe $a_{n+1} \geq aa_n$, but even this fails for similar reasons. In fact, no such inequalities can work since the left-hand side of the desired quantity equals the right-hand one when $k = \infty$. Hence the use of any intermediate inequalities cannot give the right-hand side, but rather something larger.