# Limit of a sequence proof

1. Mar 1, 2006

### r4nd0m

I am really stuck with this excercise:
Let $$a_n and b_n$$ be two sequences, where $$b_n = \frac{a_1 + a_2 + ... + a_n}{n}$$ and $$lim_(n \rightarrow \infinity) a_n = a$$ prove that $$lim_(n \rightarrow \infinity) b_n = a$$.

I tried to use the definition - $$\mid b_n - a \mid < \frac{\mid a_1 - a \mid + \mid a_2 -a \mid + ... + \mid a_n - a \mid}{n}$$, but i dont know how to proceed. Any ideas?

2. Mar 12, 2006

### AKG

Okay, try to think about what's going on. You have a sequence an going to a. So for any e > 0, you can choose a sufficiently large N such that an is within e of a, if n > N. So eventually, everything in the sequence gets as close as you want to a. Now bn is the sequence of averages. You want to show that as a sequence approaches a, so does the corresponding sequence of averages. So you want to show that for any e' > 0, you can choose N' sufficiently large so that the average bn is within e of a, for n > N'. Just choose your N' so large that regardless of the ai that are far away from a, there are enough ai, with i < N' close to a so that the average from a1 to aN' is still close to a. If you have infinitely many numbers, and "most" of them are close to a, then you can choose a finite number of them so that even of some of the ones you choose are very far from a, so many of them are close to a that the average is close to a. And of course, if the average is close for N', then it will work for all bn where n > N'. Do you get the idea? If so, then you just need to state it rigorously, and if you get the idea then it shouldn't be hard.

3. Mar 13, 2006

### benorin

It didn't display when I loaded it, so I fixed it in quote.

Last edited: Mar 13, 2006