Homework Help: 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

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