If a sequence converges to L, so the the sequence of averages

[dianetics]

Homework Statement

( x_n ) is given.

y_m = ( x₁ + . . . + x_m ) / m

show that if x_n converges to M, then y_m converges to M

Homework Equations

Definition of convergence for sequences

The Attempt at a Solution

I've tried a whole page full of algebraic manipulation. I found a recursive definition of y_m, found x_n in terms of y_m and y_m-1, and I've tried showing that abs(x_n - L) < e, implies abs(b_m - L) < e, all to no avail. I'd really appreciate some guidance.

 

[clamtrox]

Okay, so this is just a stupid idea from a physicist, but...

Suppose $$x_n$$ converges to M. Then $$z_n \equiv x_n - M$$ converges to 0, right? It would seem to me that the calculation would be a bit simpler if you worked with z instead of x.

 

[snipez90]

I think you can get some decent estimates from just algebraic manipulation, but iirc the trick here is to work with both indexes. We know that |x_n - L| < e for n > N and we need m to be sufficiently large (m > M for some M) so that |y_m - L| holds (I'm using L for the limit since I reserve M to be the measure of how large m needs to be to have convergence). Now we only care about the terms of x_n for sufficiently large n, so try working with |x_n - L| < e for n > N. Can you find an estimate involving the terms x_N, x_(N+1), ... x_(N+M)? (If this approach works as it should, you should be able to choose M at a later point)

 

[dianetics]

Can you find an estimate involving the terms x_N, x_(N+1), ... x_(N+M)?

No, lol...

 

[lanedance]

calmtrox gives a good idea, but carrying on from snipez's comments, you know for any e>0, that you can pick N so that |x_n -L|< e, for all n>N

then you know say for m > n >N that
$$(x_n+ x_{n+1}+ ...+x_m)/(m-n) \leq e$$

assume the terms are bounded below N. Then the basic idea is that you can take as many terms with n>N to smooth out any devations from the terms (as at the end of the day, everything gets divided by a big number M, that you choose)

EDIT: changed commas to plus signs

 

[dianetics]

then you know say for m > n >N that
$$(x_n, x_{n+1}, ...,x_m)/(m-n) \leq e$$

How does this imply the convergence of y_n?

 

[dianetics]

I think you can get some decent estimates from just algebraic manipulation, but iirc the trick here is to work with both indexes. We know that |x_n - L| < e for n > N and we need m to be sufficiently large (m > M for some M) so that |y_m - L| holds (I'm using L for the limit since I reserve M to be the measure of how large m needs to be to have convergence). Now we only care about the terms of x_n for sufficiently large n, so try working with |x_n - L| < e for n > N. Can you find an estimate involving the terms x_N, x_(N+1), ... x_(N+M)? (If this approach works as it should, you should be able to choose M at a later point)

I'm not following. Why do we only care about the terms of x_n for sufficiently large n? y_n is dependent on all x₁...x_n-1 as well.

Edit: I have this so far:

the convergence of x_n means that:
| (x__n + x__n+1 + ... + x_m) / (m - n + 1) - L | < e

Now I'm trying to incorporate y_m somehow into this inequality.

 

[clamtrox]

Okay, here's what I got in 15 minutes. I'm a physicist so forgive me if I'm stupid :)

Using my previous definition for z_n, you have $$|y_n - M| = (z_1 + z_2 + ... + z_n)/n$$.

Now, y_n converges if for all epsilon you can find a big enough N that for all n>N, the expression above is smaller than epsilon.

But you also know that z_n converges to zero, so you can also find a large O so that for any positive real number delta, |z_n| < delta for all n>O.

The trick with the rest of the exercise is to notice that you can always choose your delta to be smaller than epsilon. It should be very straightforward then to evaluate that

$$(z_1 + z_2 +...+ z_N + ... z_m)/m < (z_1 + z_2 + ... +z_N + \epsilon + \epsilon +...)/m$$ etc, the rest is trivial. The key part is choosing delta to be smaller than epsilon (which you can always do).

 

[lamar333]

Here's how I did it. (There are several ways, but I think this way is the most direct.)

x_n-->L & y_n=(x_1+...+x_n)/n ==> y_n-->L

PROOF:
x_n-->L, so L-e<x_n<L+e for all n>N.

So, x_n/(L-e)>1 and x_n/(L+e)<1.

Now, let y_n

=[x_1+...+x_(N-1)]/n + [x_N+...+x_n]/n.

The first term converges to 0. We focus on the second.

Note that,

(x_N+...+x_n)/n = (x_N+...+x_n)/(n-N+1) - N(x_N+...+x_n)/[n(n-N+1)] + (x_N+...+x_n)/[n(n-N+1)].

Also, by hypothesis,

x_N/(L+e)+...+x_n/(L+e) < n-N+1, since each term is less than 1.
So,
(x_N+...+x_n)/(n-N+1) < (L+e).
And,
x_N/(L-e)+...+x_n/(L-e) > n-N+1
so,
(x_N+...+x_n)/(n-N+1) > (L-e).
Thus,
(x_N+...+x_n)/(n-N+1) --> L.
Thus,
(x_N+...+x_n)/n = (x_N+...+x_n)/(n-N+1) - N(x_N+...+x_n)/[n(n-N+1)] + (x_N+...+x_n)/[n(n-N+1)] --> L - 0(L) + 0(L) = L.

Thus, y_n--> L.

 

[Rebooter]

Homework Statement

( x_n ) is given.

y_m = ( x₁ + . . . + x_m ) / m

show that if x_n converges to M, then y_m converges to M

Sorry I don't like any of the answers here so here I go:

y_m = the average of a convergent sequence.
Here's the slim jim of it all without wasting pages upon pages to prove it.
x₁...x_m approaches M when divided by m ... this becomes M! (infinite sequence approaching any number divided by infinity = that number)

ym therefore = M ... easy

 

[lanedance]

hasn't this been shut for years?

 

[lamar333]

Yeah, I was just asked to prove this using nothing but the definition of convergence. As I struggled with it, I came across this thread and decided to post what I got.