1. ## Bounded sequences.

This is an analysis question I got stuck on:

Prove that if a sequence is eventually bounded then it is bounded.
I decided to make $a_N$ my sequence. I need to show that this is bounded if $a_{N+n}$ is bounded.

So my first step was:

$|a_{n+N}|<\epsilon$

I was not sure where to go from here. Is there a way to separate $a_{n+N}$? I was also thinking that changing my epsilon to an expression would help but I can't see what to change it to.

Help would be very much appreciated.

2. You have correctly said that "eventually" means "from some point onwards", in other words for n>N. But you haven't used the definition of "bounded" (which doesn't use epsilon at all). Write down the definition of what it means for the sequence $a_{n+N}$ to be bounded, and then think how that would be affected if you include the finite number of terms $a_1,\,a_2,\ldots,\,a_N$ at the beginning of the sequence.

3. You have correctly said that "eventually" means "from some point onwards", in other words for n>N. But you haven't used the definition of "bounded" (which doesn't use epsilon at all). Write down the definition of what it means for the sequence to be bounded, and then think how that would be affected if you include the finite number of terms at the beginning of the sequence.
I was trying to use epsilon as my starting point but now I see that it was a red herring!

A sequence $(a_n)$ of real numbers is called a bounded sequence if $|a_n| \leq M$ for all n and some positive real number M. I understand what this means and i'm fine with it.

The finite number of terms at the beginning of the sequence are bounded between the maximum of $a_n$ and the minimum of $a_n$ for $n \leq N$. Therefore these finite number of terms are already bounded and it is only necessary to prove that the infinite number of terms that follow ( $|a_n| \leq M \ \forall \ n>N$) are bounded.

This is where I think the question essentially answers itself. Form the above argument, the finite number of terms at the start are bounded and the question states that the sequence is "eventually bounded" so all the infinite number of terms for n>N are also bounded. Hence the entire sequence is bounded.

Is this right?

4. Originally Posted by Showcase_22
A sequence $(a_n)$ of real numbers is called a bounded sequence if $|a_n| \leq M$ for all n and some positive real number M. I understand what this means and i'm fine with it.
To finish I suggest using $B = M + \sum\limits_{n = 1}^{N - 1} {\left| {a_n } \right|}$ as your bound.

5. Originally Posted by Showcase_22

The finite number of terms at the beginning of the sequence are bounded between the maximum of $a_n$ and the minimum of $a_n$ for $n \leq N$. Therefore these finite number of terms are already bounded and it is only necessary to prove that the infinite number of terms that follow ( $|a_n| \leq M \ \forall \ n>N$) are bounded.

This is where I think the question essentially answers itself. From the above argument, the finite number of terms at the start are bounded and the question states that the sequence is "eventually bounded" so all the infinite number of terms for n>N are also bounded. Hence the entire sequence is bounded.
So you're saying this is correct? =O

To finish I suggest using $
B = M + \sum\limits_{n = 1}^{N - 1} {\left| {a_n } \right|}
$
Could I also use $B=M+ max (a_n)$? It's not that I don't like your answer, i'm just trying to test my understanding my offering alternative solutions.

6. Originally Posted by Showcase_22
So you're saying this is correct? =O

Could I also use $B=M+ \max (a_n)$? It's not that I don't like your answer, i'm just trying to test my understanding my offering alternative solutions.
Yes, the method is correct, and the bound $B=M+ \max (a_n)$ is partially correct. You would need to replace $\max (a_n)$ by $\max (|a_n|)$, and make it clear that the "max" is meant to be taken over n=1,2,...,N. You could also use the bound $\max\{|a_1|,|a_2|,\ldots,|a_N|,M\}$.

7. Why is it necessary to include the modulus sign? In my previous post I put the modulus sign in and then changed it because I thought it wasn't needed. =S

Also, what's the lower bound? Is it just $
B=-M- \max |a_n|
$
?

8. Originally Posted by Showcase_22
Why is it necessary to include the modulus sign?
If you don't use the modulus, then you have to provide both an upper bound and a lower bound. Taking the modulus is a device for dealing with both at once, because $|a_n| means $-M.