# Thread: Limit of a recursively defined sequence

1. ## Limit of a recursively defined sequence

I'm having some trouble with this question.

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

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

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

Would I do the question like this:

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

$\displaystyle a_1=1<2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Can someone check whether what I did is correct and help me with (1) and (5)
$\displaystyle f(x)=\sqrt{1+x}$ is increasing for $\displaystyle x \in [0,\infty)$ , so we hace:

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

CB

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

Can someone check whether what I did is correct and help me with (1) and (5)
You have shown that:

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

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

CB

4. Originally Posted by CaptainBlack
You have shown that:

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

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

CB
I managed to work out (5)

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

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

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

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

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

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

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:

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

implies that (since the limit exists):

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

or:

$\displaystyle a=\sqrt{1+a}$

CB