# Sequence that converges

1. Sep 29, 2013

### Lee33

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

I was given this homework problem:

Show that if $a_1,a_2, ... ,$ is a sequence of real numbers that converges to $a$, then $$lim_{n\to \infty}\frac{\sum^n_{k=1} a_k}{n}=a.$$

I was provided a solution but my book never went over such examples or the concrete steps to solve such a problem. I am wondering what are the basic first steps to solving these types of problems?

And if possible, where can I find practice problems like these online? I searched but I couldn't find any.

Last edited: Sep 29, 2013
2. Sep 29, 2013

### arildno

Hint:
Remember that for every n, you may rewrite:
$$na=\sum_{k=1}^{n}a$$

3. Sep 29, 2013

### pasmith

There are various techniques for limit problems, but since this problem asks you to start with an arbitrary convergent sequence $(a_k)$ the only one which will work is to use what you know about $(a_k)$: for all $\epsilon > 0$ there exists $K \in \mathbb{N}$ such that if $k \geq K$ then $a - \epsilon < a_k < a + \epsilon$.

That suggests taking an arbitrary $\epsilon > 0$ and its corresponding $K$ and splitting the sum as follows:
$$\frac1n \sum_{k=1}^n a_k = \frac1n \sum_{k=1}^{K-1} a_k + \frac1n \sum_{k=K}^n a_k$$
(You are interested in the limit $n \to \infty$, so at some stage you will have $n > K$ and you may as well assume that to start with.)

Your plan is to show that
$$a - \epsilon \leq \lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k \leq a + \epsilon$$
and since $\epsilon > 0$ was arbitrary it must follow that
$$\lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k = a$$
as required.

4. Sep 29, 2013

### Staff: Mentor

I would use ϵ/2 in one step instead of ϵ, that makes the inequalities easier to show.

5. Sep 29, 2013

### Lee33

Thank you very much, pasmith! That cleared some issues I had, thanks for clarifying it for me.