# Limit of a recursively defined sequence

• September 13th 2010, 02:52 AM
acevipa
Limit of a recursively defined sequence
I'm having some trouble with this question.

Suppose that $a_1=1$ and $a_{n+1}=\sqrt{1+a_n}$ whenever $n\geq 1$

1) Show that $\sqrt{1+x}\in [1,2]$ whenever $x\in [1,2]$.

2) Use induction to show that the sequence $\{a_n\}$ is bounded.

Would I do the question like this:

Prove that $a_n<2 \ \ \ \forall n\geq 1$

$a_1=1<2$

Assume $a_k<2$, then $a_{k+1}=\sqrt{1+a_k}<\sqrt{3}<2$

By induction it follows that $a_n<2 \ \ \ \forall n\geq 1 \Rightarrow a_n$ is bounded above by 2.

3) Use induction to show that ${a_n}$ is an increasing sequence

Prove $a_{n+1}>a_n \ \ \ \forall n\geq 1$

$a_2=\sqrt{1+1}=\sqrt{2}>a_1=1$

Assume that $a_{k+1}>a_k$

$a_{k+2}=\sqrt{1+a_{k+1}}>\sqrt{1+a_k}=a_{k+1}$

By induction it follows that, $a_{n+1}>a_n \ \ \ \forall n\geq 1 \Rightarrow a_n$ is an increasing sequence

4) Explain why $\lim_{x \to \infty}a_n$ exists

Since $a_n$ is a monotonically increasing sequence, bounded above, it must converge to a limit as $n\rightarrow\infty$

5) Find $\lim_{x \to \infty}a_n$. (That is, find $\sqrt{1+\sqrt{1+\sqrt{1+...}}}$

Can someone check whether what I did is correct and help me with (1) and (5)
• September 13th 2010, 03:06 AM
CaptainBlack
Quote:

Originally Posted by acevipa
I'm having some trouble with this question.

Suppose that $a_1=1$ and $a_{n+1}=\sqrt{1+a_n}$ whenever $n\geq 1$

1) Show that $\sqrt{1+x}\in [1,2]$ whenever $x\in [1,2]$.

Can someone check whether what I did is correct and help me with (1) and (5)

$f(x)=\sqrt{1+x}$ is increasing for $x \in [0,\infty)$ , so we hace:

$1\le f(x)\le \sqrt{3}<2$ for $x\in [1,2]$

CB
• September 13th 2010, 03:08 AM
CaptainBlack
Quote:

Originally Posted by acevipa
5) Find $\lim_{x \to \infty}a_n$. (That is, find $\sqrt{1+\sqrt{1+\sqrt{1+...}}}$

Can someone check whether what I did is correct and help me with (1) and (5)

You have shown that:

$a=\lim_{x \to \infty}a_n$

exists, and so is a solution of $a=\sqrt{1+a}$

CB
• September 13th 2010, 03:12 AM
acevipa
Quote:

Originally Posted by CaptainBlack
You have shown that:

$a=\lim_{x \to \infty}a_n$

exists, and so is a solution of $a=\sqrt{1+a}$

CB

I managed to work out (5)

$\lim_{n \to \infty}a_{n+1}=\lim_{n \to \infty}\sqrt{1+a_n}\Rightarrow a=\sqrt{1+a}\Rightarrow a^2=1+a$

$\Rightarrow a^2-a-1=0\Rightarrow a=\frac{1\pm\sqrt{1+4}}{2}$

But $a_n>0$, so $\lim_{n\to\infty}a_n=\frac{1+\sqrt{5}}{2}$

By the way, is my question 2, 3, 4 correct?
• September 13th 2010, 03:33 AM
CaptainBlack
Quote:

Originally Posted by acevipa
I don't quite understand what you did.

The answer in the back of the book is $\frac{1+\sqrt{5}}{2}$

By the way, is my question 2, 3, 4 correct?

2, 3 and 4 look OK, but could use some tidying up.

what I did was to observe that:

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

implies that (since the limit exists):

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

or:

$a=\sqrt{1+a}$

CB