# Thread: Solving for the limit of this sequence

1. ## Solving for the limit of this sequence

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

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

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

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

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

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

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

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