Investigating a Sequence

**jules027** · May 17th 2008, 07:08 AM

The sequence is recursively defined by a1 = √(2), an+1 = √(2+an) for n ≥ 1.

The sequence {an} is bounded above. Call this upper bound L.

Show {an} is monotone increasing. Thus {an} has a limit. Why?

Find this limit?

Thank you sooo much for all the help guys!!!

**flyingsquirrel** · May 17th 2008, 07:56 AM

Hi

Originally Posted by jules027

The sequence is recursively defined by a1 = √(2), an+1 = √(2+an) for n ≥ 1.

The sequence {an} is bounded above. Call this upper bound L.

Show {an} is monotone increasing. Thus {an} has a limit. Why?

Let $f:x\mapsto\sqrt{2+x}$ .

$a_{n+1}=\sqrt{2+a_n}=f(a_n)$ hence the sequence is monotone increasing iff $f$ is monotone increasing.

Find this limit?

The limit $l$ of the sequence is such that $f(l)=l$ because $f$ is continuous on $[\sqrt{2},\,L]$ .

**TXGirl** · May 17th 2008, 08:01 AM

Originally Posted by jules027

The sequence {an} is bounded above. Call this upper bound L.

...{an} is monotone increasing. Thus {an} has a limit. Why?

You have by definition that a number L is the limit of a sequence $({x_n:n\in\mathbb{N}})$ iff $\forall{\epsilon >0} \exists{N}\in\mathbb{N}:\forall{n>N}, {|{x_n}-L|}<\epsilon$

A sequence converges iff it has a limit.

You can see from the definition that every monotone bounded sequence has a limit and thus converges.

So your sequence $(a_n)$ being both monotone and bounded, necessarily converges and thus has a limit.

**jules027** · May 17th 2008, 08:11 AM

Originally Posted by flyingsquirrel

Hi

Let $f:x\mapsto\sqrt{2+x}$ .

$a_{n+1}=\sqrt{2+a_n}=f(a_n)$ hence the sequence is monotone increasing iff $f$ is monotone increasing.
The limit $l$ of the sequence is such that $f(l)=l$ because $f$ is continuous on $[\sqrt{2},\,L]$ .

So there would be no computations to reach the limit? It is just the statement,
"The limit $l$ of the sequence is such that $f(l)=l$ because $f$ is continuous on $[\sqrt{2},\,L]$ ."?

**Isomorphism** · May 17th 2008, 08:14 AM

Originally Posted by jules027

So there would be no computations to reach the limit? It is just the statement,
"The limit $l$ of the sequence is such that $f(l)=l$ because $f$ is continuous on $[\sqrt{2},\,L]$ ."?

Actually $f(a_n) = a_{n+1}$ . If the limit of the sequence is L, then by letting n tend to infinity we get $f(L) = L$

$\sqrt{2+L} = L \Rightarrow L^2 - L - 2 = 0$

**flyingsquirrel** · May 17th 2008, 08:42 AM

Originally Posted by jules027

The sequence {an} is bounded above. Call this upper bound L.

Originally Posted by Isomorphism

If the limit of the sequence is L ...

These two $L$ are not necessarily the same one.

**jules027** · May 17th 2008, 03:11 PM

So how would I get at this limit exactly then? Thanks for the help!

**Isomorphism** · May 17th 2008, 08:43 PM

Originally Posted by Isomorphism

Actually $f(a_n) = a_{n+1}$ . If the limit of the sequence is L, then by letting n tend to infinity we get $f(L) = L$

$\sqrt{2+L} = L \Rightarrow L^2 - L - 2 = 0$

Originally Posted by flyingsquirrel

These two $L$ are not necessarily the same one.

flyingsquirrel means generally they need not be the same one. Since what I said could be misleading. You might think it works for any function.

However for $a_{n+1} = \sqrt{2 + a_n}$ , we can limit both sides to infinity to get $\lim_{n \to \infty} a_{n+1} =\lim_{n \to \infty} \sqrt{2 + a_n} = \sqrt{2 + \lim_{n \to \infty} a_n}$

Since you have already proved that $(a_n)$ has a limit. Call it l. Its easy to see that $\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = l$ . So you can continue like what I did before.

**jules027** · May 19th 2008, 02:51 PM

I am still confused about this one. Could somebody try and help? Maybe a step by step procedure of how to get at the answer? I almost have it but I am confused on the steps to get there. Thanks so much!!

**Mathstud28** · May 19th 2008, 03:02 PM

Originally Posted by Isomorphism

flyingsquirrel means generally they need not be the same one. Since what I said could be misleading. You might think it works for any function.

However for $a_{n+1} = \sqrt{2 + a_n}$ , we can limit both sides to infinity to get $\lim_{n \to \infty} a_{n+1} =\lim_{n \to \infty} \sqrt{2 + a_n} = \sqrt{2 + \lim_{n \to \infty} a_n}$

Since you have already proved that $(a_n)$ has a limit. Call it l. Its easy to see that $\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = l$ . So you can continue like what I did before.

As Isomorphism showed

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

and since as $n\to\infty$ $a_n=a_{n+1}$

we say Let $a_n=\lambda$

So we get

$\lambda^2=2+\lambda\Rightarrow{\lambda^2-2-\lambda=0}$

Now just solve for $\lambda$

**jules027** · May 19th 2008, 05:48 PM

So the final limit would be from (-1,2)?

**Mathstud28** · May 19th 2008, 05:51 PM

Originally Posted by jules027

So the final limit would be from (-1,2)?

No...Unless I am misreading your answer it should be just 2

Thread: Investigating a Sequence

LinkBack

Thread Tools

Display

Investigating a Sequence

Similar Math Help Forum Discussions

investigating an operator

Investigating limits of functions of several variables

investigating the second derivative test

Investigating Maths Game

investigating probabiltiy of card games

Search Tags