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Problem statement:

Determine the limit of the following sequence:

$\sqrt{a},\sqrt{1+\sqrt{a}}, \sqrt{1+\sqrt{1+\sqrt{a}}},... $

My progress:

Let´s begin by introducing some notation. Let $a_{n}$ denote the nth term of the sequence. We have $a_{1}=\sqrt{a}$ and $a_{n}=\sqrt{1+a_{n-1}}$. My instinct tells me now to rewrite as $a_{n}^2-a_{n-1}-1=0$ which has a root $\frac{1+\sqrt{5}}{2}$ (neglect the negative root for obvious reasons).

However: My friend told me this is only an eventually value of the sequence and not necessarily. I have to determine that this sequence converges before i can conclude this. How can I do this? And what does it actually mean when I solve the quadratic(because that is only an instinct of mine)?

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Correct, I´ll change that. – EricAm Dec 29 '12 at 1:20
up vote 4 down vote accepted

Your instinct was right in solving the quadratic:

If you know that the sequence is convergent with limit $l$, then making $n \to \infty$ in $a_{n}^2-a_{n-1}-1=0$ yields

$$l^2-l-1=0 \,.$$

But, as your friend told you, this helps you only if the sequence is convergent.

To prove convergence, you'll see that the root of the quadratic play a huge role. You need to distinguish among three cases:

Case 1:

$a= \frac{1+\sqrt{5}}{2}$. Then you can prove by induction that $a_n= \frac{1+\sqrt{5}}{2}$.

Case 2:

$a< \frac{1+\sqrt{5}}{2}$. Then you can prove by induction that $a_n< \frac{1+\sqrt{5}}{2}$ and $a_n <a_{n+1}$. Conclude that $a_n$ is convergent.

Case 3:

$a>\frac{1+\sqrt{5}}{2}$. Then you can prove by induction that $a_n> \frac{1+\sqrt{5}}{2}$ and $a_n >a_{n+1}$. Conclude that $a_n$ is convergent.

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Or you can just use the fact that $f(x)=\sqrt{1+x}$ is a contraction in $(1,\infty)$ and thus must have a fixed point. Specifically $|f(x)-f(y)| < \frac{1}{\sqrt{2}}|x-y|$ – Thomas Andrews Dec 29 '12 at 1:33
If $a_n\to l$ then $a_n^2\to l^2$ and $a_{n-1}\to l$ so $a_n^2-a_{n-1}-1\to l^2-l-1$ @AdamYac – Thomas Andrews Dec 29 '12 at 1:35
Thank you Thomas, I fully understand. – EricAm Dec 29 '12 at 1:39

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