# Math Help - Limit Point Question

1. ## Limit Point Question

Suppose a series $z_n \to \alpha$ as $n \to \infty$.

(That is to say $( z_n )$ has a limit point $\alpha$ at infinity.)

Prove that
$\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha$

I have found a number of series where I can prove it case by case, but am having some trouble generalizing. In addition, if the series $z_n = 1 - \frac{1}{n}$ I am having an embarrassing time showing the result at all.

Another question: Does this theorem have an actual use? I can't think of one.

(FYI: This problem is actually designed for the series to be complex. I figure if I can find out how to prove it for real series then I can generalize it to complex numbers myself.)

Thanks!
-Dan

2. Originally Posted by topsquark
Suppose a series $z_n \to \alpha$ as $n \to \infty$.

(That is to say $( z_n )$ has a limit point $\alpha$ at infinity.)

Prove that
$\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha$

I have found a number of series where I can prove it case by case, but am having some trouble generalizing. In addition, if the series $z_n = 1 - \frac{1}{n}$ I am having an embarrassing time showing the result at all.

[snip]
If $z_n = 1 - \frac{1}{n}$ then

$\frac{z_1 + z_2 + ~...~ + z_n}{n} = \frac{ \left( 1 - \frac{1}{1} \right) + \left( 1 - \frac{1}{2} \right) + \left( 1 - \frac{1}{3} \right) + ~...~ + \left( 1 - \frac{1}{n} \right)}{n}$

$= \frac{n - \sum_{k=1}^{n} \frac{1}{k}}{n}$

$= 1 - \frac{\sum_{k=1}^{n} \frac{1}{k}}{n}$.

$\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \frac{1}{k}}{n} = 0$ since the harmonic series diverges to infinity a heck of a lot slower than n does (I know that's not a solid technical argument but I'm sure it can be made so easily ..... is the glib 'easily' business the sticking point?)

3. Originally Posted by topsquark
Suppose a series $z_n \to \alpha$ as $n \to \infty$.

(That is to say $( z_n )$ has a limit point $\alpha$ at infinity.)

Prove that
$\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha$

I have found a number of series where I can prove it case by case, but am having some trouble generalizing.
[snip]
This seems way too glib to be a correct proof:

If $z_n \rightarrow \alpha$ as $n \to \infty$, then $z_1 \rightarrow \alpha$, $z_2 \rightarrow \alpha$, .....

Then $\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \lim_{n \to \infty} \frac{\alpha + \alpha + ~...~ + \alpha}{n} = \lim_{n \to \infty} \frac{n \alpha}{n} = \alpha$ ....

But like I said ....

4. Originally Posted by topsquark
Suppose a series $z_n \to \alpha$ as $n \to \infty$.

(That is to say $( z_n )$ has a limit point $\alpha$ at infinity.)
Do you really mean series or do you mean sequence?
One does not usually say that a series has a limit point.

5. I proved this for my analysis class last year . It usues real numbers, but you can easily generalize it. Here.

6. Originally Posted by Plato
Do you really mean series or do you mean sequence?
One does not usually say that a series has a limit point.
Yes, you are correct. I should have said sequence. (What can I say, it was a late night last night!)

-Dan

7. Originally Posted by mr fantastic
$\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \lim_{n \to \infty} \frac{\alpha + \alpha + ~...~ + \alpha}{n} = \lim_{n \to \infty} \frac{n \alpha}{n} = \alpha$ ....
Thank you for the effort, but this statement is incorrect. Perhaps the mistake in terminology Plato pointed out is a problem. So let me restate the original problem in more correct terms:

Suppose a sequence $( z_n )$ has the property $z_n \to \alpha$ as $n \to \infty$.

(That is to say $( z_n )$ has a limit point $\alpha$ at infinity.)

Prove that
$\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha$

Since we are talking about a sequence here we cannot say that $z_n \to \alpha$ for all n, we are saying that the sequence approaches a limit point $\alpha$, as n approaches infinity. That is to say there is only one sequence, not an infinite number of sequences tending toward the same limit.

-Dan

8. Originally Posted by ThePerfectHacker
I proved this for my analysis class last year . It usues real numbers, but you can easily generalize it. Here.
I am somewhat confused almost from the beginning by the line
$| s _n - s | = \left | \frac{s_1 + ~...~ + s_n - s \cdot n}{n} \right |$

Is this again a case of my mistaken wording? If you define $s_n$ to be the nth partial sum of a series, this statement is quite understandable and falls out naturally. (Hopefully by now it is clear that this is not what I meant.)

-Dan

9. Originally Posted by topsquark
Suppose a series $z_n \to \alpha$ as $n \to \infty$.

(That is to say $( z_n )$ has a limit point $\alpha$ at infinity.)

Prove that
$\lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha$
There's a neat proof of this by CaptainBlack here for the case where α=0. You can easily deduce the general case from this by applying it to the sequence $(z_n-\alpha)$.

Originally Posted by topsquark
Another question: Does this theorem have an actual use? I can't think of one.
This is a case of Cesàro summation, which is an essential tool in Fourier theory. The main result about convergence of Fourier series (Fejér's theorem) is proved by using Cesàro summation.

So yes, the fact that ordinary convergence implies Cesàro convergence is certainly useful.

10. Topsquark try proving this similar result: Let $x_n>0$ and $\lim ~ x_n = x$. Prove that the geometric means $(x_1...x_k)^{1/k} \to x$.

11. Originally Posted by topsquark
I am somewhat confused almost from the beginning by the line
$| s _n - s | = \left | \frac{s_1 + ~...~ + s_n - s \cdot n}{n} \right |$

Is this again a case of my mistaken wording? If you define $s_n$ to be the nth partial sum of a series, this statement is quite understandable and falls out naturally. (Hopefully by now it is clear that this is not what I meant.)

-Dan
I am using $\sigma_n$ not $s_n$. Where $\sigma_n$ is the sequence of averages.