Cauchy Sequences and Convergence

[Seth|MMORSE]

Homework Statement

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.

Homework 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

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?

 

[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.

 

[Seth|MMORSE]

Case 1: (a_{n}) is an increasing sequence
\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a&lt;\frac{n*a}{n}-a=a-a=0<br /> \Rightarrow b_{n}-a&lt;0

Case 2: (a_{n}) is a decreasing sequence
\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a&gt;\frac{n*a}{n}-a=a-a=0<br /> \Rightarrow b_{n}-a&gt;0

\Rightarrow 0&lt;b_{n}-a&lt;0 <br /> <br /> \Rightarrow |b_{n}-a|&lt;0<br /> <br /> \Rightarrow b_{n}\rightarrow a

Does this make any sense?

 

[LCKurtz]

Case 1: (a_{n}) is an increasing sequence
\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a&lt;\frac{n*a}{n}-a=a-a=0<br /> \Rightarrow b_{n}-a&lt;0

Case 2: (a_{n}) is a decreasing sequence
\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a&gt;\frac{n*a}{n}-a=a-a=0<br /> \Rightarrow b_{n}-a&gt;0

\Rightarrow 0&lt;b_{n}-a&lt;0 <br /> <br /> \Rightarrow |b_{n}-a|&lt;0<br /> <br /> \Rightarrow b_{n}\rightarrow a

Does this make any sense?

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.

 

[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 $a_n\rightarrow a$.

 

[Seth|MMORSE]

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 $a_n\rightarrow a$.

Okay, thanks! Will try ...

 

[Seth|MMORSE]

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 $a_n\rightarrow a$.

a_{n}\rightarrow a \Rightarrow |a_{n}-a|&lt;\epsilon, \forall n&gt;N

Assume |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|&lt;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| &lt; 2\epsilon-\epsilon
\Rightarrow |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a_{n}|&lt;\epsilon

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

 

[LCKurtz]

a_{n}\rightarrow a means that

Given $\epsilon > 0$ there exists $N$ such that

|a_{n}-a|&lt;\epsilon, \forall n&gt;N

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.

Assume |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|&lt;2\epsilon

But that is basically what you are trying to prove. You can't start by assuming it is true.

|\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| &lt; 2\epsilon-\epsilon
\Rightarrow |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a_{n}|&lt;\epsilon

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

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.

 

[Seth|MMORSE]

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.

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]
&lt;\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|&lt;\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 ...

 

[LCKurtz]

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

$|b_{n}-a|=|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|$
$=|\frac{(a_{1}+a_{2}+...+a_{n})-na}{n}|$
$=|\frac{((a_{1}-a)+(a_{2}-a)+...+(a_{n}-a))}{n}|$
$\leq |\frac{(a_1-a)}{n}|+|\frac{(a_2-a)}{n}|+...+|\frac{(a_n-a)}{n}|$

Obviously missing out on something (or totally off track?) since there was nothing relevant to your last sentence.

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.

 

[Seth|MMORSE]

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.

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?

 

[LCKurtz]

Homework Statement

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.

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 ...

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 $\epsilon$.

I don't get it

What end result am I looking for?

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.

 

[Seth|MMORSE]

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.

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.

 

[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.

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.

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

 

[Seth|MMORSE]

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

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

 

[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 ...

 

[LCKurtz]

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 ...

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

 

[Seth|MMORSE]

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

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

 

[LCKurtz]

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

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

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.

 

[Seth|MMORSE]

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.

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

 

[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.

 

[Seth|MMORSE]

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.

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 $n>N_0, (1-\frac{N_0}{n})<1$
$(1-\frac{N_0}{n})(\frac{\epsilon}{2})<\frac{\epsilon}{2}$
$\Rightarrow q<\frac{\epsilon}{2}$

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

$∴(b_n)\rightarrow a$

 

[LCKurtz]

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 $\epsilon>0, \exists n\in \mathbb{N}:|a_n-a|<\epsilon\ \forall n>N$

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 $\epsilon>0, \exists N\in \mathbb{N}:|a_n-a|<\epsilon\ \forall n>N$

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

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 $\epsilon>0, \exists N\in \mathbb{N}:|a_n-a|<\epsilon/2\ \forall n>N$

Show $|b_n-a|<\epsilon$

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.

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

$p=|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|$
$=\frac{1}{n}(|\frac{a_1-a}{n}|+...+|\frac{a_{N_0}-a}{n}|)$

Check your arithmetic there and fix the 3 lines below.

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 $n>N_0, (1-\frac{N_0}{n})<1$
$(1-\frac{N_0}{n})(\frac{\epsilon}{2})<\frac{\epsilon}{2}$
$\Rightarrow q<\frac{\epsilon}{2}$

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

$∴(b_n)\rightarrow a$

You aren't done until you give me the value of $M$ that works. One more iteration should fix your argument.

 

[Seth|MMORSE]

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.

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 ...

 

[LCKurtz]

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

Much clearer to let $N_1=\frac{2(|a_1-a|+...+|a_{N}-a|)}{\epsilon}$ and say if $n>N_1$ then $p<\frac{\epsilon}{2}$

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

$|b_n-a|=p+q$

Shouldn't that be $\le$?

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

$M=N?$

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.

 

[Seth|MMORSE]

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.

Does it have anything to do with the fact that I used $N$ for the first sequence and I used $\frac{\epsilon}{2}$ as a bound?

 

[LCKurtz]

Does it have anything to do with the fact that I used $N$ for the first sequence and I used $\frac{\epsilon}{2}$ as a bound?

Only in the fact that if both pieces are less than $\frac{\epsilon}{2}$ their sum will be less than $\epsilon$. How big does $n$ have to be to get both parts to work?

 

[Seth|MMORSE]

Only in the fact that if both pieces are less than $\frac{\epsilon}{2}$ their sum will be less than $\epsilon$. How big does $n$ have to be to get both parts to work?

Like the one stated in the workings? $n>$ than something for the first section and just $>N$ for the second one.

I can visualise $a_n$ eventually lying within $\epsilon/2$'s after $N$, but how would I know where M would be for $b_n$?

 

[LCKurtz]

Like the one stated in the workings? $n>$ than something for the first section and just $>N$ for the second one.

I can visualise $a_n$ eventually lying within $\epsilon/2$'s after $N$, but how would I know where M would be for $b_n$?

We have already talked about that. Look back and absorb what we have talked about.

 

[Seth|MMORSE]

We have already talked about that. Look back and absorb what we have talked about.

$M>N_1$ that you defined? I have the feeling that I have to start learning analysis from scratch again ...

 

[LCKurtz]

$M>N_1$ that you defined? I have the feeling that I have to start learning analysis from scratch again ...

Maybe so. This is like pulling teeth. I have already given more help than I should on this forum.

That will make $p<\epsilon/2$. You need both p and q to be less than $\epsilon/2$. How big does M need to be so that if $n>M$ they both work? You should be able to figure that out.

 

[Seth|MMORSE]

Maybe so. This is like pulling teeth. I have already given more help than I should on this forum.

That will make $p<\epsilon/2$. You need both p and q to be less than $\epsilon/2$. How big does M need to be so that if $n>M$ they both work? You should be able to figure that out.

Since for $n>N$, we are able to get $|b_n-a|<\epsilon$ ...
Therefore, if $n>M\geq N, |b_n-a|<\epsilon$

 

[LCKurtz]

Since for $n>N$, we are able to get $|b_n-a|<\epsilon$ ...
Therefore, if $n>M\geq N$, $|b_n-a|<\epsilon$

But maybe $M$ isn't greater than $N$.
[Edit] I am not going to be satisfied with your answer until you have something like:

Let M = ______. Then if $n>M$ both p and q are less then $\epsilon/2$.

 

[Seth|MMORSE]

But maybe $M$ isn't greater than $N$.

Therefore it should be there exists an $M\geq N$ such that $|b_n-a|<\epsilon\ \forall n>M$?

 

[LCKurtz]

I edited post #33.

 

[Seth|MMORSE]

$M=\frac{a_1+a_2+...+a_M}{\epsilon+a}$?

From equating $\epsilon$ and $b_M-a$

 

[LCKurtz]

No. The best I can do without giving you the answer, which I'm not going to do, is tell you to re-read post #25. You are stuck on a really easy issue, unfortunately.

 

[Seth|MMORSE]

No. The best I can do without giving you the answer, which I'm not going to do, is tell you to re-read post #25. You are stuck on a really easy issue, unfortunately.

I don't see anything in post #25
I'm can't help it that my brain doesn't know what to look for, just staring at it over and over again without seeing anything different ... sigh

 

[LCKurtz]

I don't see anything in post #25
I'm can't help it that my brain doesn't know what to look for, just staring at it over and over again without seeing anything different ... sigh

In post #24 you had, among other things:

$|b_n-a|$
$= |\frac{a_1+a_2+...+a_n}{n}-a|$
$=|\frac{a_1+a_2+...+a_n-na}{n}|$
$=|\frac{(a_1-a)+(a_2-a)+...+(a_n-1)}{n}|$
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 $n>\frac{2(|a_1-a|+...+|a_{N}-a|)}{\epsilon}$
$\frac{1}{n}(|a_1-a|+...+|a_{N}-a|)<\frac{\epsilon}{2}$
$\Rightarrow p<\frac{\epsilon}{2}$

To which I responded in post #25:

Much clearer to let $N_1=\frac{2(|a_1-a|+...+|a_{N}-a|)}{\epsilon}$ and say:
if $n>N_1$ then $p<\frac{\epsilon}{2}$

You also had in post #24 this:

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

So if $n>N$ then $q<\frac{\epsilon}{2}$

Then you concluded that

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

That conclusion is only valid if both p and q are less than $\epsilon/2$. Look at the statements I highlighted in red. How large does M need to be to make both statements true if $n>M$?

 

[Seth|MMORSE]

$M=max(N,N_1)$?
I kept thinking of using inequalities ...

 

[LCKurtz]

$M=max(N,N_1)$?
I kept thinking of using inequalities ...

There now. That wasn't so hard, was it? I hope you have learned a few things about $\epsilon,N$ proofs in this thread. So you will just breeze through your next such problem.

 

[Seth|MMORSE]

There now. That wasn't so hard, was it? I hope you have learned a few things about $\epsilon,N$ proofs in this thread. So you will just breeze through your next such problem.

NONE of my previous exercises required me to use such a function ... never occurred to me ... till something struck.

Anyway, thank you so much for your help!