1. Increasing monotonic series

Hey.

I have recursive series:

$\forall n \in \mathbb{N}, \ \ a_{n+1}=\sqrt{2+a_n}, \ \ a_1=t, \ \ -2\leq t \in\mathbb{R}$

Prove:

If $-2\leq t\leq 2$ then A(n) is defined to every natural n and the series A(n) is an increasing monotonic series and is bounded above.

We have studied the material so far up to Cantor's lemma, but I don't know how to start here.

I have begun by trying to prove that this series is at least an increasing-monotonic but I got that this inequality has to be true:
$a_n^2<2+a_n$
Which I don't know how to confirm.

As to bounded above, have no idea how to think here on bounds...

Thanks.

2. Re: Increasing monotonic series

Originally Posted by CStudent
I have recursive series:
$\forall n \in \mathbb{N}, \ \ a_{n+1}=\sqrt{2+a_n}, \ \ a_1=t, \ \ -2\leq t \in\mathbb{R}$

Prove:
If $-2\leq t\leq 2$ then A(n) is defined to every natural n and the series A(n) is an increasing monotonic series and is bounded above.
There are mistakes in the statement. You write about a series and then proceed to define a sequence $a_n$. THEN you proceed further to ask about a function $A(n)$ without defining it.
Is $\displaystyle A(n) = \sum\limits_{k = 1}^n {{a_k}}$ or is it simply $A(n)=a_n~?$ Is it a series or is it a sequence?

3. Re: Increasing monotonic series

first prove by induction that

$\displaystyle a_n\leq 2$

4. Re: Increasing monotonic series

Originally Posted by Plato
There are mistakes in the statement. You write about a series and then proceed to define a sequence $a_n$. THEN you proceed further to ask about a function $A(n)$ without defining it.
Is $\displaystyle A(n) = \sum\limits_{k = 1}^n {{a_k}}$ or is it simply $A(n)=a_n~?$ Is it a series or is it a sequence?
It's a series

5. Re: Increasing monotonic series

Originally Posted by Idea
first prove by induction that

$\displaystyle a_n\leq 2$
How do you realize at first glance that this series is bounded above by 2?

Thanks.

6. Re: Increasing monotonic series

Originally Posted by CStudent
How do you realize at first glance that this series is bounded above by 2?

Thanks.
we want to show that $\displaystyle a_n^2 <2+a_n$

but $\displaystyle x^2<2+x$ for $\displaystyle x<2$

so we must prove $\displaystyle a_n<2$ and then

$\displaystyle a_n^2 <2a_n<a_n+2$ and so

$\displaystyle a_n<\sqrt{2+a_n}=a_{n+1}$