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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 $b_i\in \mathbb Q,(i=0,1,2,\cdots)$?

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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.

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  • $\begingroup$ What is $b_0$? $b_0=1$? $\endgroup$
    – lsr314
    Jul 28 '13 at 3:49
  • $\begingroup$ Yes, I guess it is going to be $1$. $\endgroup$
    – OR.
    Jul 28 '13 at 3:51
  • $\begingroup$ What is $a_1$? $a_1$ is the initial term of $A_n,$ it must be a fixed number. $\endgroup$
    – lsr314
    Jul 28 '13 at 3:56
  • $\begingroup$ The sequence $b_n$ determines it. In this case $a_1=b_1^2=(1-1/2)^2=1/4$. $\endgroup$
    – OR.
    Jul 28 '13 at 3:59
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    $\begingroup$ $a_1$ is only determined by $b_1$. From the formula for $b_n$, we have $b_n,a_1,\ldots,a_{n-1}$ already defined so we solve for $a_n$. $\endgroup$
    – OR.
    Jul 28 '13 at 4:18

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