# Simple sequence proof on general metric space

1. Sep 21, 2010

### tjackson3

1. The problem statement, all variables and given/known data

Consider the sequence $$a_1,a_2,...$$, such that $$\lim_{n\rightarrow\infty} a_n = a$$ (with $$a_i \in$$ R). Show that $$\lim_{n\rightarrow\infty}\left(\frac{\sum_{i=1}^n a_i}{n}\right) = a$$

In other words, it's given that for some $$\epsilon > 0,d(a_n,a) < \epsilon\ \forall n > N$$

2. Relevant equations

The limit of a sequence is defined as follows: Let $$\epsilon > 0$$. The limit of a sequence $$a_1,a_2,...$$ is a if and only if for all n > N, $$d(a_n,a) < \epsilon$$.

Possibly useful too (as it came up in one of my thoughts on this proof) is the definition of a Cauchy sequence. Note that any convergent sequence on any metric space is a Cauchy sequence, and a Cauchy sequence is defined as follows: Let $$\epsilon > 0$$. A sequence is a Cauchy sequence if and only if for every n > N, $$d(a_n,a_{n+1}) < \epsilon$$.

3. The attempt at a solution

I have so many scribbled attempts at solutions on a couple of sheets of scratch paper here that it's hard to put them down coherently. Here are the ideas I've had so far:

1.) We can split up the fraction to get that

$$\lim_{n\rightarrow\infty}\left(\frac{a_1}{n} + \frac{a_2}{n} + ...\right) = a$$

This implies that

$$\lim_{n\rightarrow\infty}\frac{a_1}{n} + \lim_{n\rightarrow\infty}\frac{a_2}{n} + ... = a$$

Where to go from there I'm not sure. My instinct was to collect all terms for n > N, use the fact that this sequence was Cauchy, and say that

$$\frac{a_1}{n} + \frac{a_2}{n} + ... + \frac{a_N}{n} + \frac{a_{N+1}}{n} + ... < \frac{a_1}{n} + \frac{a_2}{n} + ... + \frac{(n-N)a_N}{n}$$

which would simplify things, but we can't assume that (since this assumes that the sequence is decreasing; alternatively, flipping the inequality assumes that the sequence is increasing).

2.) Multiply both sides of the limit by n. Then we get $$\lim_{n\rightarrow\infty}\sum_{i=1}^n a_i = \lim_{n\rightarrow\infty}an$$, which means that it would be necessary to prove that $$d(\lim_{n\rightarrow\infty}\sum_{i=1}^n a_i, \lim_{n\rightarrow\infty}an) < \epsilon$$. My gut instinct on where to go from here is to find some intermediate quantity to use with the triangle inequality, but I don't know where to get that from.

3.) From 2, subtract an from both sides to get $$\lim_{n\rightarrow\infty}\left(\sum_{i=1}^n a_i - an\right) = 0$$, but again, I don't know where to go from there aside from possibly using the triangle inequality.

4.) One other thing I had considered, relating to #1 above, is that for n > N, write $$a_n = a_{N+1} + \epsilon_n$$, where $$\epsilon_n = d(a_n,a)$$. However, I think that this method might imply the use of the Euclidean metric, but I'm not sure. I suppose that I could use a sandwich theorem with $$a_n = a_{N+1} \pm \epsilon_n$$, but again, that wouldn't help if I'm implicitly using the Euclidean metric.

Thanks so much for your help!

2. Sep 21, 2010

### Office_Shredder

Staff Emeritus
A couple of your ideas aren't strictly possible. For example, once you multiply both sides by n, you have two limits that don't exist anymore. And when you try to split up the limit, you can't do this since the number of terms in the summand changes as n changes; it's not a finite number of functions being added together, just one function changing really.

Here's the basic idea:

If $$|a_i-a|<\epsilon$$ for $$M<n$$ say, then pick some huge $$N>>M$$ and consider $$\frac{\sum_{i=1}^N a_i}{N}$$. Split this up into two parts: The first M terms, and the rest. The rest will sum to something very close to a (since each individual ai is close to a), and the first M terms, if N is large enough, will be negligible. Basically, you have to use the values that you know are good for the limit as a starting point for values that will be good for the summation, and then find another value that is larger enough to make it all work

3. Sep 21, 2010

### tjackson3

Office_Shredder: Thank you! I suppose my problem is making it rigorous. So splitting it up into two parts gives:

$$\frac{\sum_{i=1}^N a_i}{N} = \frac{\sum_{i=1}^M a_i}{M} + \frac{\sum_{i=M+1}^N a_i}{N}$$

I don't understand how to show how either of the two sums achieves the value that it should in a general metric space. For example, how would one show that $$d(\frac{\sum_{i=1}^M a_i}{M},0) < \epsilon$$ for $$M>N$$?

4. Sep 21, 2010

### Office_Shredder

Staff Emeritus
What do you mean by a general metric space? You mean a general choice of metric on the real numbers?

When you split it up you did it wrong. It should be
$$\frac{\sum_{i=1}^N a_i}{N} = \frac{\sum_{i=1}^M a_i}{N} + \frac{\sum_{i=M+1}^N a_i}{N}$$

With an N in the denominator of the first summand. This is important because $$\frac{\sum_{i=1}^M a_i}{N}$$ obviously becomes arbitrarily small as N becomes big once you've fixed an M.

$$\frac{\sum_{i=M+1}^N a_i}{N}$$ once you've fixed M will not converge to a. Instead, you want to prove that it can get to within a given $$\epsilon$$ by making N big. Then if you want to get closer, pick a different M and re-find what N should be.

So to re-iterate the strategy:
If you want to show that you can get within $$\epsilon$$ of a, you need to pick a value of M so that $$|a_i-a|$$ is small enough (this might not be $$<\epsilon$$, you might want to pick a smaller number to give yourself room for error at the end). Then find a value of N to make the summation get within $$\epsilon$$ of a.

5. Sep 21, 2010

### tjackson3

Sorry, having M in the denominator there was a typo. The issue is really that I feel like I can make $$|a_i - a|$$ small enough due to algebraic manipulation. The problem is getting $$d(a_i,a)$$ small.