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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.2012-12-29

1 Answers 1

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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 . 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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    Thank you Thomas, I fully understand.2012-12-29