Cauchy Sequences and Convergence

[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!