Epsilon-limit proof for real number sequences

[kingwinner]

Homework Statement

Let {a_n} be a sequence of real numbers.
Suppose a_n->L as n->∞. Prove that [(a₁+a₂+...+a_n)/n] ->L as n->∞.

Homework Equations

N/A

The Attempt at a Solution

By definition:
a_n->L iff
for all ε>0, there exists an integer N such that n≥N => |a_n - L|< ε.

Given ε>0.
I start with |[(a₁+a₂+...+a_n)/n] - L|. But I cannot think of a way to link this with |a_n - L| so I have no idea how to continue.

Any help is appreciated! :)

 

[kingwinner]

I think we may have to use the triangle inequality, but still I have no idea how to get into the exact the same form as |a_n - L|.

Could someone help me, please?

 

[kingwinner]

Given \varepsilon &gt; 0 there is N such that if n > N, then you can make

|\frac{a_1 + ... + a_n}{n}| &lt; \frac \varepsilon 4

I don't get this part.

In the hypothesis, we suppose a_n->L, so:
for all ε_a>0, there exists an integer N such that n≥N => |a_n - L|< ε_a.

So for example:
Given ε>0. We can find N s.t. n≥N => |a_n-L|< ε/4

But how come you have |(a1+...+an) / n|?

Thanks for your help! :)

 

[LCKurtz]

I don't get this part.

In the hypothesis, we suppose a_n->L, so:
for all ε_a>0, there exists an integer N such that n≥N => |a_n - L|< ε_a.

So for example:
Given ε>0. We can find N s.t. n≥N => |a_n-L|< ε/4

But how come you have |(a1+...+an) / n|?

Thanks for your help! :)

It isn't. I must have been in a fog when I posted that and I have deleted it. Be back a little later.

 

[LCKurtz]

Given \varepsilon &gt; 0 pick N such that if n > N then |a_n - L| &lt; \varepsilon/ 4. Now if n > N

\left|\frac{a_1+...+a_n}{n}-L\right| = \left|\frac{a_1+...+a_n}{n}-\frac {nL}{n}\right|= \left|\frac{(a_1-L)+...+(a_n-L)} n\right|

= \left| \frac{(a_1-L)+...+(a_N-L)}{n}+\frac{(a_{N+1}-L)+...+(a_n-L)}{n}\right|

Now use the triangle inequality. Since N is fixed you can make the first term small in n is large enough. Then work on the second term to make it small. Sorry for the false start earlier.

 

[kingwinner]

Given \varepsilon &gt; 0 pick N such that if n > N then |a_n - L| &lt; \varepsilon/ 4. Now if n > N

\left|\frac{a_1+...+a_n}{n}-L\right| = \left|\frac{a_1+...+a_n}{n}-\frac {nL}{n}\right|= \left|\frac{(a_1-L)+...+(a_n-L)} n\right|

= \left| \frac{(a_1-L)+...+(a_N-L)}{n}+\frac{(a_{N+1}-L)+...+(a_n-L)}{n}\right|

Now use the triangle inequality. Since N is fixed you can make the first term small in n is large enough. Then work on the second term to make it small. Sorry for the false start earlier.

1) hmm...why did you choose ε/4 in |a_n - L| &lt; \varepsilon/ 4??

2) At the end, we have to show that:
for all ε>0, there exists an integer M such that
n≥M => |[(a₁+a₂+...+a_n)/n] - L|< ε.
But how can we find such an M?

Thanks for your help! :)

 

[LCKurtz]

hmm...why did you choose ε/4 in |a_n - L| &lt; \varepsilon/ 4??

Because I anticipate adding a couple of small terms together and having the sum come out less than \varepsilon.

At the end, we have to show that:
for all ε>0, there exists an integer M such that
n≥M => |[(a₁+a₂+...+a_n)/n] - L|< ε.
But how can we find such an M?

Thanks for your help! :)

Have you done the triangle inequality on this yet, so you have two terms to make small? If you can say there exists an N₁ such that n > N₁ makes the first term small and an N₂ such that n > N₂ makes the second term small, then if you take N_max = max{N₁,N₂,N}, wouldn't that do it? Of course, you have to fill in the arguments why you can pick these N's.

 

[kingwinner]

Because I anticipate adding a couple of small terms together and having the sum come out less than \varepsilon.



Have you done the triangle inequality on this yet, so you have two terms to make small? If you can say there exists an N₁ such that n > N₁ makes the first term small and an N₂ such that n > N₂ makes the second term small, then if you take N_max = max{N₁,N₂,N}, wouldn't that do it? Of course, you have to fill in the arguments why you can pick these N's.

So after using the triangle inequality (which will break it into 2 terms), for the first term I have to find N₁ such that n≥N₁ => |[(a₁ -L)+...(a_N -L)]/n| < ε/2

To find this N₁, I think I have to relate |[(a₁ -L)+...(a_N -L)]/n| to |a_n - L|, but how is this possible?? I can't think of a way of doing it...

Thanks!

 

[LCKurtz]

N is a fixed, finite, number. There are finitely many terms in the numerator and you have n in the denominator. Just pick n big enough...

 

[kingwinner]

N is a fixed, finite, number. There are finitely many terms in the numerator and you have n in the denominator. Just pick n big enough...

I get the intuitive idea, but just like every epsilon-limit proof of this type, we have to actually find a specific N₁ which works, right? And I'm stuck on this step...

 

[LCKurtz]

I get the intuitive idea, but just like every epsilon-limit proof of this type, we have to actually find a specific N₁ which works, right? And I'm stuck on this step...

So solve your inequality |[(a1 -L)+...(aN -L)]/n| < ε/2 for n to see how large it must be.

 

[kingwinner]

So solve your inequality |[(a1 -L)+...(aN -L)]/n| < ε/2 for n to see how large it must be.

OK. I think it is n> 2 |(a₁-L)+...+(a_N-L)| / ε.
So I think we can take N₁ = 2 |(a₁-L)+...+(a_N-L)| / ε, is this correct?
But on top of N depending on ε, N also depends on the terms in the sequence, is this OK? (i.e. is this allowed?)

 

[LCKurtz]

OK. I think it is n> 2 |(a₁-L)+...+(a_N-L)| / ε.
So I think we can take N₁ = 2 |(a₁-L)+...+(a_N-L)| / ε, is this correct?
But on top of N depending on ε, N also depends on the terms in the sequence, is this OK? (i.e. is this allowed?)

Everything depends on the terms of the sequence. You wouldn't expect the same N to work for all sequences, would you?

Remember you might have to take N₁ slightly larger than that because that might not be an integer.

 

[kingwinner]

So I guess I will take N₁ to be the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.


Any hints for the second term? (sorry I'm asking it but I really have no clue...)

|[a_N+1 - L +...+ (a_n-1-L) +(a_n-L)] / n|

Here there are terms like a_n, a_n-1, etc. The "n" pops up everyhwere, so I don't think I can possibly solve for n in the inequality just like I did before.

 

[r.a.c.]

The sequence an converges therefore it is bounded. So every an<M where M is some number. So the sum becomes less that (M-L) ...n-N+1 times over n and from there go on.

 

[kingwinner]

The sequence an converges therefore it is bounded. So every an<M where M is some number. So the sum becomes less that (M-L) ...n-N+1 times over n and from there go on.

|[a_N+1 - L +...+ (a_n-1-L) +(a_n-L)] / n|
≤|(n-N)(M-L)/n|

I think it should be n-N instead of n-N+1?
n-(N+1)+1 = n-N

 

[LCKurtz]

So I guess I will take N₁ to be the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.


Any hints for the second term? (sorry I'm asking it but I really have no clue...)

|[a_N+1 - L +...+ (a_n-1-L) +(a_n-L)] / n|

Here there are terms like a_n, a_n-1, etc. The "n" pops up everyhwere, so I don't think I can possibly solve for n in the inequality just like I did before.

Somewhere in this proof you need to use the fact that the original sequence converges to L. Remember at the beginning where you had that if n > N,

<br /> |a_n - L| &lt; \varepsilon/ 4<br />

Use that now.

 

[kingwinner]

Somewhere in this proof you need to use the fact that the original sequence converges to L. Remember at the beginning where you had that if n > N,

<br /> |a_n - L| &lt; \varepsilon/ 4<br />

Use that now.

Do I have to use the fact that any convergence sequence is bounded?

But how should I handle the terms like (a_n-1-L) in the numerator?

thanks.

 

[LCKurtz]

Do I have to use the fact that any convergence sequence is bounded?

But how should I handle the terms like (a_n-1-L) in the numerator?

thanks.

I fear you are getting lost in the subscripts. In the expression:

|[a_N+1 - L +...+ (a_n-1-L) +(a_n-L)] / n|

all the subscripts are greater than N. So every one of them satisfies the inequality:

<br /> <br /> |a_k - L| &lt; \varepsilon/ 4<br /> <br />

for whatever value from N+1 to n that k takes. Use that.

 

[kingwinner]

I fear you are getting lost in the subscripts. In the expression:

|[a_N+1 - L +...+ (a_n-1-L) +(a_n-L)] / n|

all the subscripts are greater than N. So every one of them satisfies the inequality:

<br /> <br /> |a_k - L| &lt; \varepsilon/ 4<br /> <br />

for whatever value from N+1 to n that k takes. Use that.

So should I break |[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| using the triangle inequality into n-N terms?
|[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| <= (1/n)[|a_N+1 - L| + ...+ |a_n-1-L| + |a_n-L|] ?

But I don't think it will work...

 

[LCKurtz]

So should I break |[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| using the triangle inequality into n-N terms?
|[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| <= (1/n)[|a_N+1 - L| + ...+ |a_n-1-L| + |a_n-L|] ?

But I don't think it will work...

You won't know until you try. How big are the terms? How many terms are there?

 

[kingwinner]

You won't know until you try. How big are the terms? How many terms are there?

There are n-N terms.

|[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| <= (1/n)[|a_N+1 - L| + ...+ |a_n-1-L| + |a_n-L|] < (1/n)(n-N)ε/4

I can't solve for n here as I did for the first part, so I am not sure what to do next...

 

[LCKurtz]

How big can

\left(\frac {n-N}{n}\right)\left(\frac\varepsilon 4\right)

be?

 

[kingwinner]

How big can

\left(\frac {n-N}{n}\right)\left(\frac\varepsilon 4\right)

be?

It goes to 0 as n->infinity.

But how can we find a specific N₂ which works (just like before when we found N₁) in the epsilon-limit proof?

 

[LCKurtz]

You aren't looking for another N and we aren't talking about n -> infinity. All you are trying to do is show this sum is small. You have that it is smaller than

<br /> \left(\frac {n-N}{n}\right)\left(\frac\varepsilon 4\right)<br />

I will ask you again. Is that small? Look at it and think about it.

 

[kingwinner]

You aren't looking for another N and we aren't talking about n -> infinity. All you are trying to do is show this sum is small. You have that it is smaller than

<br /> \left(\frac {n-N}{n}\right)\left(\frac\varepsilon 4\right)<br />

I will ask you again. Is that small? Look at it and think about it.

I think we can say it's less than ε/4?

 

[LCKurtz]

You think?? Are you sure? If the answer is yes, then you should be able to look back at what has been done and see how to determine an N such the if n > N then

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| &lt; \varepsilon<br />

At least I hope so. Sack time here.

Edit: This morning I noticed a typo, meant \varepsilon above, not \varepsilon/4

 

[kingwinner]

Yes, I pretty sure it's less than or equal to epsilon/4, because N is some positive integer.
So we've have |(a_1 +...+a_n)/n|< ε/2 + ε/4 = 3ε/4 < ε

If you can say there exists an N₁ such that n > N₁ makes the first term small and an N₂ such that n > N₂ makes the second term small, then if you take N_max = max{N₁,N₂,N}, wouldn't that do it?

But I am a bit confused now. According to what you said above (in the quote), we should be looking for a workable N₂. How can we find it?

thanks!

 

[LCKurtz]

You just need to review what has been done, keep the various N's straight, and understand it. Given that a_n converges to L, you are trying to show that

\frac {a_1 + ... + a_n} n \rightarrow L.

So this means that given \varepsilon &gt; 0, you must find an integer N such that if n > N then

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| &lt; \varepsilon<br />

As in post #5, you know you can pick N₁ such that if n > N₁ then |a_n - L| &lt; \varepsilon/ 4

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| \leq \left| \frac{(a_1-L)+...+(a_{N_1}-L)}{n}+\frac{(a_{N_1+1}-L)+...+(a_n-L)}{n}\right|<br />

And you applied the triangle inequality to that to get:

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| \leq \left| \frac{(a_1-L)+...+(a_{N_1}-L)}{n}\right|+\left|\frac{(a_{N_1+1}-L)+...+(a_n-L)}{n}\right|<br />

Now, in post #14 you saw now big you had to have n to be to make the first term small. Now let's call that N by the name N₂, since we are already using N₁ and want to save N for the final one. And remember that if N₂ isn't already greater than N₁, take it large enough so that it is, so now if n > N₂ you have the first term less than \varepsilon/2 and |a_n - L| &lt; \varepsilon/ 4.

Now use what you know about the second term and you should be able to see the finish line.

 

[kingwinner]

You just need to review what has been done, keep the various N's straight, and understand it. Given that a_n converges to L, you are trying to show that

\frac {a_1 + ... + a_n} n \rightarrow L.

So this means that given \varepsilon &gt; 0, you must find an integer N such that if n > N then

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| &lt; \varepsilon<br />

As in post #5, you know you can pick N₁ such that if n > N₁ then |a_n - L| &lt; \varepsilon/ 4

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| \leq \left| \frac{(a_1-L)+...+(a_{N_1}-L)}{n}+\frac{(a_{N_1+1}-L)+...+(a_n-L)}{n}\right|<br />

And you applied the triangle inequality to that to get:

<br /> \left|\frac{a_1+...+a_n}{n}-L\right| \leq \left| \frac{(a_1-L)+...+(a_{N_1}-L)}{n}\right|+\left|\frac{(a_{N_1+1}-L)+...+(a_n-L)}{n}\right|<br />

Now, in post #14 you saw now big you had to have n to be to make the first term small. Now let's call that N by the name N₂, since we are already using N₁ and want to save N for the final one. And remember that if N₂ isn't already greater than N₁, take it large enough so that it is, so now if n > N₂ you have the first term less than \varepsilon/2 and |a_n - L| &lt; \varepsilon/ 4.

Now use what you know about the second term and you should be able to see the finish line.

I think this would work.
Take N=max{N₁,N₂}
where N₁ is as you defined
and N₂ = the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.
If n > N, then |[(a₁+a₂+...+a_n)/n] - L|< ε.


However, according to what you said in post #7, we should take N to be the max of THREE things, i.e. N=max{N₁,N₂,N₃}. Why? What is N₃? I don't understand where this third restriction comes from.
I believe |[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| <= (1/n)[|a_N+1 - L| + ...+ |a_n-1-L| + |a_n-L|] < (1/n)(n-N)ε/4 < ε/4 is always true, isn't it? So what is N₃?

Thanks for your time and patience! (this topic is really new to me)

 

[LCKurtz]

I think this would work.
Take N=max{N₁,N₂}
where N₁ is as you defined
and N₂ = the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.
If n > N, then |[(a₁+a₂+...+a_n)/n] - L|< ε.

That is correct.

However, according to what you said in post #7, we should take N to be the max of THREE things, i.e. N=max{N₁,N₂,N₃}. Why? What is N₃? I don't understand where this third restriction comes from.
I believe |[(a_N+1 - L) +...+ (a_n-1-L) +(a_n-L)] / n| <= (1/n)[|a_N+1 - L| + ...+ |a_n-1-L| + |a_n-L|] < (1/n)(n-N)ε/4 < ε/4 is always true, isn't it? So what is N₃?

In our back-and-forth we may have called N different things; that's why I summarized it in post #29. We don't need an N₃.