# Homework Help: Analysis - Cauchy sequences?

1. Nov 7, 2012

### Seth|MMORSE

1. The problem statement, all variables and given/known data
Prove the following theorem, originally due to Cauchy. Suppose that $(a_{n})$$\rightarrow a$. Then the sequence $(b_{n})$ defined by $b_{n}=\frac{(a_{1}+a_{2}+...+a_{n})}{n}$ is convergent and $(b_{n})$$\rightarrow a$.

2. Relevant equations
A sequence $(a_{n})$ has the Cauchy property if, for each $ε>0$ there exists a natural number N such that $|a_{n} - a_{m}|<ε$ for all $n,m>N$

3. The attempt at a solution
I don't exactly know what am I suppose to do here ... ?
The only thing I could think of is $(a_{1}+a_{2}+...+a_{n})≈n*a_{n}$ for a very huge n but that doesn't seem really relevant.

Am I suppose to show that $(b_{n})$ is a subsequence? Or $(b_{n})$ is somewhat related to the $(a_{m})$ in the definition of the Cauchy sequence?

Last edited: Nov 7, 2012
2. Nov 7, 2012

### LCKurtz

You know $a_n\rightarrow a$ and you want to make$$\left| \frac{a_1+a_2+...+a_n}{n}-a\right |$$small if $n$ is large enough. Start with that.

3. Nov 7, 2012

### Seth|MMORSE

Case 1: $(a_{n})$ is an increasing sequence
$\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a<\frac{n*a}{n}-a=a-a=0 \Rightarrow b_{n}-a<0$

Case 2: $(a_{n})$ is a decreasing sequence
$\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a>\frac{n*a}{n}-a=a-a=0 \Rightarrow b_{n}-a>0$

$\Rightarrow 0<b_{n}-a<0 \Rightarrow |b_{n}-a|<0 \Rightarrow b_{n}\rightarrow a$

Does this make any sense?

4. Nov 7, 2012

### LCKurtz

I'm afraid not. The sequence can't be both increasing and decreasing unless it is a constant, and it may not be either one anyway. I would try to find an $\epsilon,\, N$ type of argument.

5. Nov 7, 2012

### LCKurtz

I have to leave for a couple of hours. Let me just add not to waste your time using the Cauchy criteria. All you need is the $\epsilon,\, N$ definition of $a_n\rightarrow a$.

6. Nov 7, 2012

### Seth|MMORSE

Okay, thanks! Will try ...

7. Nov 8, 2012

### Seth|MMORSE

$a_{n}\rightarrow a \Rightarrow |a_{n}-a|<\epsilon, \forall n>N$

Assume $|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|<2\epsilon$

$|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a-(a_{n}-a)| \leq |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|-|a_{n}-a| < 2\epsilon-\epsilon$
$\Rightarrow |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a_{n}|<\epsilon$

Okay I honestly have no idea where am I going with this.

8. Nov 8, 2012

### LCKurtz

Given $\epsilon > 0$ there exists $N$ such that
If you are going to do an $\epsilon,\, N$ proof, it is best to give complete and exact statements. And it might be a good idea to use $\epsilon/2$ or $\epsilon/4$ to give yourself some extra room.
But that is basically what you are trying to prove. You can't start by assuming it is true.

Yes, I see that. As I said earlier, you want to start with $$\left| \frac{a_1+a_2+...+a_n}{n}-a\right |$$
and make it small. To get started write it as a single fraction$$\left| \frac{a_1+a_2+...+a_n-na}{n}\right |$$Now since, at least for large $n$, you can make $|a_n-a|$ small, try spreading the n a's with each of the $a_i$'s. Then think about the part where $n\le N$ and where $n>N$ and see if you can make them both small for $n$ large enough.

9. Nov 8, 2012

### Seth|MMORSE

Hmm ... guess I should refrain from rephrasing definitions so carelessly.

$|b_{n}-a|=|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|$
$=|\frac{(a_{1}+a_{2}+...+a_{n})-na}{n}|$
$=|\frac{((a_{1}-a)+(a_{2}-a)+...+(a_{n}-a))}{n}|$
$\leq |\frac{(a_1-a)}{n}|+|\frac{(a_2-a)}{n}|+...+|\frac{(a_n-a)}{n}|$ [by Triangle Inequality]
$<\frac{\epsilon_1}{n}+\frac{\epsilon_2}{n}+...+$$\frac{\epsilon_{n}}{n}$ [Since $|a_n-a|<\epsilon$ and the theorem that states that subsequences (i.e. $a_1, a_2, ...$) tend to the same limit (single term subsequences as well??)]
$=\frac{n\epsilon}{n}$
$=\epsilon$
$\Rightarrow |b_{n}-a|<\epsilon$

Obviously missing out on something (or totally off track?) since there was nothing relevant to your last sentence. I understand that by definition of convergent sequences, it will only be between both $\epsilon$ if $n$ is larger than a specific point $N$ that corresponds to the $\epsilon$ chosen but ... then what?
I feel very bad for having the lack of capability to comprehend your hints ...

10. Nov 8, 2012

### LCKurtz

OK so far. Just do what I suggested. If $n>N$ part of that sum has $n\le N$ and part has $n>N$. Break it into those two parts. Then what you have to figure out is why both parts are small if $n$ is large enough (and how large "large enough" is). That is the meat of the argument.

11. Nov 8, 2012

### Seth|MMORSE

I guess for $n\le N$, this part would be small enough because of a large denominator, i.e. $n$? Whereas for $n>N, (a_n-a)$ would be small ... but I'm pretty sure this isn't what you're alluding to ...

I don't get it

What end result am I looking for? Something like the one I posted before but with different intermediate steps?

12. Nov 8, 2012

### LCKurtz

That is the idea but you must write it mathematically. And you haven't yet broken it into two parts depending on $N$. You are going to need an $N$ in there do do that. Then show how big and why that big that $n$ must be to make each of those less than $\epsilon/2$ to get the whole thing less than $\epsilon$.
What are you looking for?? If you don't know that no wonder you are confused. Read your statement in red above. And $b_n\rightarrow a$ means given $\epsilon > 0$ blah blah. You really need to learn carefully to write down what you are given and what you are trying to prove, using the definitions.

13. Nov 8, 2012

### Seth|MMORSE

Something like this??
$|\frac{(a_1-a)}{n}|+|\frac{(a_2-a)}{n}|+...+|\frac{(a_N-a)}{n}|+|\frac{(a_{N+1}-a)}{n}|+...+|\frac{(a_n-a)}{n}|$
$(|\frac{(a_1-a)}{n}|+...+|\frac{(a_N-a)}{n}|)+(|\frac{(a_{N+1}-a)}{n}|+...+|\frac{(a_n-a)}{n}|)$

To clarify, I wasn't unsure of the main objective of the question, just wanted to know whether am I suppose to get something something $<\epsilon$, to which you have answered.

14. Nov 8, 2012

### LCKurtz

That's better. Now look again at the advice from post #6 I have highlighted above.

15. Nov 9, 2012

### Seth|MMORSE

Only thing I could come up with is this:
$(|\frac{(a_1-a)}{n}|+...+|\frac{(a_N-a)}{n}|)<\frac{|a_n-a|}{2}<\frac{\epsilon}{2}$
$\frac{N(a_N-a)}{n}<\frac{|a_n-a|}{2}$
$\frac{N}{n}<\frac{1}{2}$
$n>2N$

16. Nov 9, 2012

### Seth|MMORSE

Okay, thought about it again (okay not really, kinda discussed with my friends) and this is the result.

Let $\epsilon>0, |a_n-a|<\frac{\epsilon}{2} \ \forall n>N$
$|\frac{a_{N+1}-a}{n}|+...+|\frac{a_{n}-a}{n}|<\frac{\epsilon_{N+1}}{2n}+...+\frac{ε_{n}}{2n}=(\frac{n-N}{n})(\frac{\epsilon}{2})$
Since $n>N, (\frac{n-N}{n})<1$
$\Rightarrow|\frac{a_{N+1}-a}{n}|+...+|\frac{a_{n}-a}{n}|<\frac{\epsilon}{2}$

For the remaining part ...
$|\frac{a_{1}-a}{n}|+...+|\frac{a_{N}-a}{n}|=\frac{1}{n}(|a_1-a|+...+|a_N-a|)$
$(|a_1-a|+...+|a_N-a|)$ is already fixed, therefore for a large enough $n, \frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}$

I still don't see how can I show how big of an $n$ is required ...

17. Nov 9, 2012

### LCKurtz

Solve that inequality for n to see how large it must be.

18. Nov 9, 2012

### Seth|MMORSE

$\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}$ this?
$\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}$
$\frac{2(|a_1-a|+...+|a_N-a|)}{\epsilon}<n$

19. Nov 9, 2012

### LCKurtz

I didn't notice when I scanned your post a minute ago. Where did all these $\epsilon_n$'s come from? You can't introduce new variables in a proof without saying what they represent.

20. Nov 9, 2012

### Seth|MMORSE

Oh, they're just $\epsilon$'s that correspond to each term ... which are essentially the usual $\epsilon$'s. Sorry about that.

21. Nov 9, 2012

### LCKurtz

OK, after all this discussion, let's see a complete, properly written argument, starting with what you are given, what you have to show and showing it.

22. Nov 9, 2012

### Seth|MMORSE

Suppose $(a_n)\rightarrow a$.
Show that $(b_n)\rightarrow a$ when $b_n=\frac{a_1+a_2+...+a_n}{n}$

Definition of $(a_n)\rightarrow a$:
For each $\epsilon>0, \exists n\in \mathbb{N}:|a_n-a|<\epsilon\ \forall n>N$

Let $|a_n-a|<\frac{\epsilon}{2}\ \forall n>N_0$
Show $|b_n-a|<\epsilon$

$|b_n-a|$
$= |\frac{a_1+a_2+...+a_n}{n}-a|$
$=|\frac{a_1+a_2+...+a_n-na}{n}|$
$=|\frac{(a_1-a)+(a_2-a)+...+(a_n-1)}{n}|$
By triangle inequality,
$\leq |\frac{a_1-a}{n}|+|\frac{a_2-a}{n}|+...+|\frac{a_n-a}{n}|$
$=\underbrace{(|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|)}_{p}+ \underbrace{(|\frac{a_{N_0+1}-a}{n}|+...+|\frac{a_n-a}{n}|)}_{q}$

$p=|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|$
$=\frac{1}{n}(|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|)$
We define $n>\frac{2|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|}{\epsilon}$
$\frac{1}{n}(|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|)<\frac{\epsilon}{2}$
$\Rightarrow p<\frac{\epsilon}{2}$

$q=|\frac{a_{N_0+1}-a}{n}|+...+|\frac{a_n-a}{n}|$
Using $|a_n-a|<\frac{\epsilon}{2}\ \forall n>N_0$
$q<\frac{\epsilon/2}{n}+...+\frac{\epsilon/2}{n}$
$=(\frac{n-N_0}{n})(\frac{\epsilon}{2})$
$=(1-\frac{N_0}{n})(\frac{\epsilon}{2}$
Since $n>N_0, (1-\frac{N_0}{n})<1$
$(1-\frac{N_0}{n})(\frac{\epsilon}{2})<\frac{\epsilon}{2}$
$\Rightarrow q<\frac{\epsilon}{2}$

$|b_n-a|=p+q$
$|b_n-a|< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
$|b_n-a|<\epsilon$

$∴(b_n)\rightarrow a$

23. Nov 9, 2012

### LCKurtz

You have there exists n, then you mention N, then you mention $N_0$ below, but you haven't told us what $N$ or $N_0$ is. I suppose you meant this: For each $\epsilon>0, \exists N\in \mathbb{N}:|a_n-a|<\epsilon\ \forall n>N$
You mean there exists $N_0$ such that $|a_n-a|<\frac{\epsilon}{2}\ \forall n>N_0$. But you don't need another N. Just use $\epsilon/2$ in the first place: For each $\epsilon>0, \exists N\in \mathbb{N}:|a_n-a|<\epsilon/2\ \forall n>N$

NO!! You must show there exists an $M$ such that if $n>M$ then $|b_n-a|<\epsilon$. Definitions, definitions! Everything below is dedicated to finding such an $M$, isn't it? And you can use $N$ instead of $N_0$ below as I indicated above.
Check your arithmetic there and fix the 3 lines below.
You aren't done until you give me the value of $M$ that works. One more iteration should fix your argument.

24. Nov 9, 2012

### Seth|MMORSE

Suppose $(a_n)\rightarrow a$.
Show that $(b_n)\rightarrow a$ when $b_n=\frac{a_1+a_2+...+a_n}{n}$

Definition of $(a_n)\rightarrow a$:
For each $\epsilon>0, \exists N\in \mathbb{N}:|a_n-a|<\frac{\epsilon}{2} \forall n>N$
Show $\exists M\in \mathbb{N}:|b_n-a|<\epsilon\ \forall n>M$

$|b_n-a|$
$= |\frac{a_1+a_2+...+a_n}{n}-a|$
$=|\frac{a_1+a_2+...+a_n-na}{n}|$
$=|\frac{(a_1-a)+(a_2-a)+...+(a_n-1)}{n}|$
By triangle inequality,
$\leq |\frac{a_1-a}{n}|+|\frac{a_2-a}{n}|+...+|\frac{a_n-a}{n}|$
$=\underbrace{(|\frac{a_1-a}{n}|+...+|\frac{a_{N}-a}{n}|)}_{p}+ \underbrace{(|\frac{a_{N+1}-a}{n}|+...+|\frac{a_n-a}{n}|)}_{q}$

$p=|\frac{a_1-a}{n}|+...+|\frac{a_{N}-a}{n}|$
$=\frac{1}{n}(|a_1-a|+...+|a_{N}-a|)$
We define $n>\frac{2(|a_1-a|+...+|a_{N}-a|)}{\epsilon}$
$\frac{1}{n}(|a_1-a|+...+|a_{N}-a|)<\frac{\epsilon}{2}$
$\Rightarrow p<\frac{\epsilon}{2}$

$q=|\frac{a_{N+1}-a}{n}|+...+|\frac{a_n-a}{n}|$
Using $|a_n-a|<\frac{\epsilon}{2}\ \forall n>N$
$q<\frac{\epsilon/2}{n}+...+\frac{\epsilon/2}{n}$
$=(\frac{n-N}{n})(\frac{\epsilon}{2})$
$=(1-\frac{N}{n})(\frac{\epsilon}{2})$
Since $n>N, (1-\frac{N}{n})<1$
$(1-\frac{N}{n})(\frac{\epsilon}{2})<\frac{\epsilon}{2}$
$\Rightarrow q<\frac{\epsilon}{2}$

$|b_n-a|=p+q$
$|b_n-a|< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
$|b_n-a|<\epsilon$

$M=N???$
$∴\exists M:|b_n-a|<\epsilon\ \forall n>M$
$∴(b_n)\rightarrow a$

Okay ... I guess I have to work on using definitions properly ...

25. Nov 9, 2012

### LCKurtz

Much clearer to let $N_1=\frac{2(|a_1-a|+...+|a_{N}-a|)}{\epsilon}$ and say if $n>N_1$ then $p<\frac{\epsilon}{2}$
Shouldn't that be $\le$?
No. You don't get to just pull that out of a hat like magic. Look at the argument and figure out how big M has to be to make it all work.