Investigating a Sequence

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May 17th 2008, 07:08 AM
jules027

Investigating a Sequence

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!!!
May 17th 2008, 07:56 AM
flyingsquirrel

Hi

Quote:

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.

Quote:

Find this limit?

The limit $l$ of the sequence is such that $f(l)=l$ because $f$ is continuous on $[\sqrt{2},\,L]$ .
May 17th 2008, 08:01 AM
TXGirl

Quote:

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.
May 17th 2008, 08:11 AM
jules027

Quote:

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 $[\sqrt{2},\,L]$ ."?
May 17th 2008, 08:14 AM
Isomorphism

Quote:

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 $f(L) = L$

$\sqrt{2+L} = L \Rightarrow L^2 - L - 2 = 0$
May 17th 2008, 08:42 AM
flyingsquirrel

Quote:

Originally Posted by jules027

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

Quote:

Originally Posted by Isomorphism

If the limit of the sequence is L ...

These two $L$ are not necessarily the same one.
May 17th 2008, 03:11 PM
jules027

So how would I get at this limit exactly then? Thanks for the help!
May 17th 2008, 08:43 PM
Isomorphism

Quote:

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 $f(L) = L$

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

Quote:

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.
May 19th 2008, 02:51 PM
jules027

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!!
May 19th 2008, 03:02 PM
Mathstud28

Quote:

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 $\lambda$
May 19th 2008, 05:48 PM
jules027

So the final limit would be from (-1,2)?
May 19th 2008, 05:51 PM
Mathstud28

Quote:

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