Q:

A sequence of rationals whose continued square roots are also rational

For a given sequence $A_n=\{a_1,a_2,\cdots\}$, let 
$$b_1=\sqrt{a_1},b_2=\sqrt{a_1+\sqrt{a_2}},\cdots,b_k=\sqrt{a_1+\sqrt{a_2+\sqrt{\cdots+\sqrt{a_k}}}},\cdots,b_0=\lim_{n\to \infty}b_n.$$
Do there exist a sequence $A_n$ such that $a_i\in \mathbb Q^{+},(i=1,2,\cdots)$ and $b_i\in \mathbb Q,(i=0,1,2,\cdots)$?

A:

If we define first the sequence $b_n$ this determines the sequence $a_n$. If $b_n$ is rational then $a_n$ will be rational and $\geq0$. What we need is to ensure that $a_n\neq0$. We can ensure this by asking that the denominator in $b_n$, written as a reduced fraction is divisible by a prime that doesn't divide any of the denominators of the previous $b_k$, $k<n$. So, we just need a convergent sequence with this property. For example $b_n=1-1/p_n$ where $p_n$ is the $n$-th prime.

