Epsilon-limit proof for real number sequences

**kingwinner** · January 18th 2010, 03:38 PM

[note: also under discussion in s.o.s. math board]

**Plato** · January 18th 2010, 04:08 PM

This is very well known question. It is generally known as the Sequence of Means.

It is done in two cases. First prove it for the case $L=0$ .

Then if $L\ne 0$ define $b_n=a_n-L$ and use the first case.

**kingwinner** · January 18th 2010, 04:27 PM

I don't see how to work out the epsilon-limit proof.
At the end, we have to show that:
for all ε>0, there exists an integer M such that
n≥M => |[(a1+a2+...+an)/n] - L|< ε.

But how can we find that M?

(even for the case L=0 I am facing the same trouble...I can't find a way to link |[(a1+a2+...+an)/n] - L| with |a_n -L|, so I don't know how to find M)

Any help is appreciated!

**Plato** · January 18th 2010, 04:38 PM

Have you had that “intensive one-on-one instruction/tutorial” that I suggested for you earlier?
I seems to me that you have some real issues with this material.

You need help beyond which you can expect from websites.

**kingwinner** · January 18th 2010, 05:01 PM

Originally Posted by Plato

Have you had that “intensive one-on-one instruction/tutorial” that I suggested for you earlier?
I seems to me that you have some real issues with this material.

You need help beyond which you can expect from websites.

I would LOVE to have it, but unfortunately I don't have the fortune to have such a treasury experience...I'm learning it from my textbooks.

But could you kindly point out which part I was wrong that leads you to make the above comment? I don't think I've said anything conceptually wrong yet...
We have to show that (by definition of convergence):
for all ε>0, there exists an integer M such that
n≥M => |[(a1+a2+...+an)/n] - L|< ε.

and just like every epsilon-limit proof of this type, we have to actually find a specific M that works, right? And to do this, I believe we need to link |[(a1+a2+...+an)/n] - L| with |a_n -L|. And right now, I am stuck on this step. I know that the triangle inequality may help, but I am not sure how to apply it in this case.

Could someone help me, please?
Any help is much appreciated!

**Jhevon** · January 18th 2010, 05:16 PM

Originally Posted by kingwinner

I would LOVE to have it, but I don't have the fortune to have such a treasury experience...I'm learning it from my textbooks.

But could you kindly point out which part I was wrong that leads you to make the above comment? I don't think I've said anything conceptually wrong yet...
We have to show that (by definition of convergence):
for all ε>0, there exists an integer M such that
n≥M => |[(a1+a2+...+an)/n] - L|< ε.

and just like every epsilon-limit proof of this type, we have to actually find a specific M that works, right? And to do this, I believe we need to link |[(a1+a2+...+an)/n] - L| with |a_n -L|. And right now, I am stuck on this step. I know that the triangle inequality may help, but I am not sure how to apply it in this case.

Could someone help me, please?
Any help is much appreciated!

Hey king!

How about combining the fractions?

$\frac {a_1 + a_2 + \cdots + a_n}n - L = \frac {a_1 + a_2 + \cdots + a_n - nL}n = \frac {a_1 - L + a_2 - L + \cdots a_n - L}n$ $= \frac {a_1 - L}n + \frac {a_2 - L}n + \cdots + \frac {a_n - L}n$

see how to apply the triangle inequality now?

**kingwinner** · January 18th 2010, 05:46 PM

Originally Posted by Jhevon

Hey king!

How about combining the fractions?

$\frac {a_1 + a_2 + \cdots + a_n}n - L = \frac {a_1 + a_2 + \cdots + a_n - nL}n = \frac {a_1 - L + a_2 - L + \cdots a_n - L}n$ $= \frac {a_1 - L}n + \frac {a_2 - L}n + \cdots + \frac {a_n - L}n$

see how to apply the triangle inequality now?

OK, after using the triangle inequality, I will have n terms.
|(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|

How can I bound each of these?
My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n?
I have no clue how to continue, can you please show me how?

Thanks!

**Drexel28** · January 18th 2010, 05:48 PM

Originally Posted by kingwinner

OK, after using the triangle inequality, I will have n terms.
|(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|

How can I bound each of these?
My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n.
I have no clue how to continue, can you please show me how?

Thanks!

Your $\varepsilon$ cannot depend on $n$ .

**Jhevon** · January 19th 2010, 09:45 AM

Originally Posted by kingwinner

OK, after using the triangle inequality, I will have n terms.
|(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|

How can I bound each of these?
My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n?
I have no clue how to continue, can you please show me how?

Thanks!

By the way, I obliged you, but i wouldn't really use the triangle inequality here, well, not like this at least. finding something to bound each of these expressions seems like too much work, or at least odd. pick your $M$ to be one of the subscripts in $a_1,~a_2,\dots,~a_n$ . Bound the sum of all $|a_i - L|/n$ of everything below it (that is, the expressions for which $i \le M$ ), and each term above it could be bounded by some $\epsilon$ . See if you can make something like that work.

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