# Prove every convergent sequence of real numbers is bounded &

1. Apr 29, 2013

### Zondrina

1. The problem statement, all variables and given/known data

The question : http://gyazo.com/7eb4b86c61150e4af092b9f8afeaf169

2. Relevant equations

Sup/Inf axioms
Methods of constructing sequences
$ε-N$
$lim(a_n) ≤ sup_n a_n$ from question 5 right before it.

I'll split the question into two parts.

3. The attempt at a solution

(a) We must prove every convergent sequence of real numbers is bounded. Suppose $a_n$ is a convergent sequence, that is $lim(a_n) = L$ where $L \in ℝ$.

So for $ε = 1, \exists N \space | \space n ≥ N \Rightarrow |a_n - L| < 1$

Thus $-1 + L < a_n < 1 + L$ so that $a_n$ is bounded below by -1 + L and above by 1 + L.

(b) Suppose $lim(a_n) = L$, we must prove that $inf_n a_n ≤ L ≤ sup_n a_n$

From question 5 prior, we already know that $L ≤ sup_n a_n$

So we want to show $inf_n a_n ≤ L$.

Assume the converse is true, that is suppose that $inf_n a_n > L$.

Since $inf_n a_n ≤ sup_n a_n$, we have that $sup_n a_n ≥ inf_n a_n > L$, but this implies that $sup_n a_n > L$ which is a contradiction.

2. Apr 29, 2013

### christoff

Part a)
This does not in general work. All you know is that for $n\geq N$, this bound is valid. What about for $n<N$?

3. Apr 29, 2013

### xaos

notice that a sequence {a[n]} is convergent if lim sup a[n] = lim inf a[n] = L. the idea is that as 'n' goes to infinity, the largest or smallest value in the 'tail' of the sequence approaches L monotonically [why?]. so if "inf a[n] > L" for some value of 'n', if must be true for all larger values of 'n'.

4. Apr 29, 2013

### LCKurtz

And for (b), what is the contradiction? There is no problem with $\sup a_n > L$. Also, to make your posts more readable, use a backslash in front of keywords like inf, sup, lim.

5. Apr 29, 2013

### Zondrina

I'll address this first.

After some thought I believe i see what you're intending. I don't need to consider n<N. I just need to choose a different N, say $N'$

(a) We must prove every convergent sequence of real numbers is bounded. Suppose $a_n$ is a convergent sequence, that is $\lim(a_n) = L$ where $L \in ℝ$.

So for $ε = 1, \exists N \space | \space n ≥ N \Rightarrow |a_n - L| < 1$

Thus $-1 + L < a_n < 1 + L$ so that $a_n$ is bounded below by -1 + L and above by 1 + L. This covers all cases of $n ≥ N$.

Now, for some $n < N$ we can see that $a_n > L + 1$ so that $L + 1$ is not an upper bound so we define :

$M = max\{ max\{a_n \space | \space n < N\}, L + 1 \}$ so we could bound $a_n$ as desired.

(b) Suppose $\lim(a_n) = L$, we must prove that $\inf_n a_n ≤ L ≤ \sup_n a_n$

From question 5 prior, we already know that $L ≤ \sup_n a_n$

So we want to show $\inf_n a_n ≤ L$.

Assume the converse is true, that is suppose that $\inf_n a_n > L$.

We know $\forall ε > 0, \exists N \space | \space n ≥ N \Rightarrow |a_n - L| < ε$

So for $ε = \inf_n a_n - L$ we have $|L - a_n| < \inf_n a_n - L$

So we have $-\inf_n a_n + L < a_n - L < \inf_n a_n - L$ which yields $a_n < \inf_n a_n$ which is a contradiction.

$∴ \inf_n a_n ≤ \lim(a_n) ≤ \sup_n a_n$ as desired.

OUT OF QUESTION : Is $\sup_n a_n = limsup(a_n)$? Ditto for liminf?

Last edited: Apr 30, 2013
6. Apr 29, 2013

### Dick

Well, you took that hint completely the wrong way. I don't see what N' is buying you. You've already shown $a_n$ is bounded for n>N. All you have to worry about is n<=N. That isn't so hard to deal with is it? There are only a finite number of terms.

7. Apr 29, 2013

### Zondrina

Hmm you're right, now that I look back I see I didn't really need N', but why is my $M$ not good?

I'm not quite sure why I have to consider everything below a certain number when I'm considering all terms of the sequence as well as $ε + |L|$ for some positive $ε$.

Also I added another try at part (b).

EDIT : I see I don't need my $N_2$ either now.

EDIT 2 : Wow facepalm, i just realized what n<N meant. I fixed my M accordingly.

EDIT 3 : Fixed all the errors now hopefully.

Last edited: Apr 29, 2013
8. Apr 29, 2013

### Dick

It's really hard to tell what you changed. It's still all a mess. Look, your initial try at part a) was almost correct. Just say what you are going to do with n<=N. You can't take the max over all $a_n$ until you shown it has a max. Otherwise you could have done that to begin with.

Last edited: Apr 29, 2013
9. Apr 29, 2013

### Zondrina

I dropped the unneeded N's and cleaned it up a bit.

Also my initial try? As in the one in my first post? I don't see how my most recent attempt is wrong considering I took $M = \{|a_n|, 1 + |L| ...$ ( I wrapped $a_n$ in an abs ). This means that I consider all values of $n \in \mathbb{N}$, not just values $n ≥ N$ ( I understand your issue is that I need to show $a_n$ is bounded for all n ). So I'm not convinced this is incorrect.

Considering my initial try was correct though... let me try to reason out what I would do for n < N ( chris also mentioned this in an earlier post ).

Hmm for some $n < N$ we can see that $a_n > L + 1$ so that $L + 1$ is not an upper bound so I would define :

$M = max\{ max\{a_n \space | \space n < N\}, L + 1 \}$ so we could bound $a_n$ as desired.

I still don't see what's wrong with what I mentioned prior though.

10. Apr 29, 2013

### Dick

max of $a_n$ for n<N is well defined regardless of whether the series even converges or not. You are taking the max of a FINITE number of terms. That's why it's well defined. max $a_n$ over all n is not. Until you say why it is. Your latest attempt makes that clear.

11. Apr 29, 2013

### Zondrina

Ahh yes I understand your concern now. So I can't just use the max over what may potentially be an infinite number of terms as... that's simply broken.

EDIT : Cleaned up the proof in post #5, should be good now.

Last edited: Apr 30, 2013
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