# Math Help - Solving for the limit of this sequence

1. ## Solving for the limit of this sequence

$a_{1}=1$
$a_{n+1}=\sqrt{1+\sqrt{a_{n}}}$

So, taking the limit of both sides, call the limit A, gives us
A=lim $\sqrt{1+\sqrt{a_{n}}}$= $\sqrt{1+\sqrt{A}}$

Solving for A gives us
$A^2=1+\sqrt{A}$ or
$A^2-\sqrt{A}-1=0$

I'm assuming I did all of the above correctly. My only issue now is... how do I solve for A?

2. Originally Posted by paupsers
$a_{1}=1$
$a_{n+1}=\sqrt{1+\sqrt{a_{n}}}$

So, taking the limit of both sides, call the limit A, gives us
A=lim $\sqrt{1+\sqrt{a_{n}}}$= $\sqrt{1+\sqrt{A}}$

Solving for A gives us
$A^2=1+\sqrt{A}$ or
$A^2-\sqrt{A}-1=0$

I'm assuming I did all of the above correctly. My only issue now is... how do I solve for A?
$A^2 - 1 = \sqrt{A} \Rightarrow (A^2 - 1)^2 = A \Rightarrow A^4 - 2A^2 - A + 1 = 0$. Solve for A.

There are some other things you need to do as well, like showing that the sequence is bounded.