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 $\varepsilon > 0$ there is N such that if n > N, then you can make

$$|\frac{a_1 + ... + a_n}{n}| < \frac \varepsilon 4$$

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 $\varepsilon > 0$ pick N such that if n > N then $|a_n - L| < \varepsilon/ 4$. 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 $\varepsilon > 0$ pick N such that if n > N then $|a_n - L| < \varepsilon/ 4$. 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 $|a_n - L| < \varepsilon/ 4$??

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 $|a_n - L| < \varepsilon/ 4$??

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,

$|a_n - L| < \varepsilon/ 4$

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,

$|a_n - L| < \varepsilon/ 4$

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:

$|a_k - L| < \varepsilon/ 4$

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:

$|a_k - L| < \varepsilon/ 4$

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

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

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

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

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

$$\left|\frac{a_1+...+a_n}{n}-L\right| < \varepsilon$$

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 $\varepsilon > 0$, you must find an integer N such that if n > N then

$$\left|\frac{a_1+...+a_n}{n}-L\right| < \varepsilon$$

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

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

And you applied the triangle inequality to that to get:

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

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 $|a_n - L| < \varepsilon/ 4$.

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 $\varepsilon > 0$, you must find an integer N such that if n > N then

$$\left|\frac{a_1+...+a_n}{n}-L\right| < \varepsilon$$

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

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

And you applied the triangle inequality to that to get:

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

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 $|a_n - L| < \varepsilon/ 4$.

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.