(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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]

2. Relevant 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].

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

[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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Simple sequence proof on general metric space

**Physics Forums | Science Articles, Homework Help, Discussion**