- #1

tjackson3

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## Homework Statement

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

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

## Homework Equations

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

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 [tex]\epsilon > 0[/tex]. A sequence is a Cauchy sequence if and only if for every n > N, [tex]d(a_n,a_{n+1}) < \epsilon[/tex].

## 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

[tex]\lim_{n\rightarrow\infty}\left(\frac{a_1}{n} + \frac{a_2}{n} + ...\right) = a[/tex]

This implies that

[tex]\lim_{n\rightarrow\infty}\frac{a_1}{n} + \lim_{n\rightarrow\infty}\frac{a_2}{n} + ... = a[/tex]

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

[tex]\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}[/tex]

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 [tex]\lim_{n\rightarrow\infty}\sum_{i=1}^n a_i = \lim_{n\rightarrow\infty}an[/tex], which means that it would be necessary to prove that [tex]d(\lim_{n\rightarrow\infty}\sum_{i=1}^n a_i, \lim_{n\rightarrow\infty}an) < \epsilon[/tex]. 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 [tex]\lim_{n\rightarrow\infty}\left(\sum_{i=1}^n a_i - an\right) = 0[/tex], 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 [tex]a_n = a_{N+1} + \epsilon_n[/tex], where [tex]\epsilon_n = d(a_n,a)[/tex]. 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 [tex]a_n = a_{N+1} \pm \epsilon_n[/tex], but again, that wouldn't help if I'm implicitly using the Euclidean metric.

Thanks so much for your help!