How to Bound the Limit of Arithmetic Mean from Below

[JG89]

Homework Statement

Prove if that if the limit of a_n = c as n approaches infinity, then the limit of o_n = c as n approaches infinity, where o_n is the arithmetic mean (a_1 + ... + a_n)/n

Homework Equations

I can't figure out how to bound it from below.

The Attempt at a Solution

Assuming that the terms go in ascending order, then a_n is the largest term in the numerator, and so o_n <= n(a_n)/n = a_n.

So I have it bounded above. But I can't figure out how to bound it below, such that the sequence which bounds it below converges to a_n as well. Help would be appreciated!

 

[statdad]

Try this. Select \epsilon &gt; 0. Since the original sequence converges to c there is an integer N_1 such that

<br /> |a_n - c | &lt; \epsilon \quad \text{ for } n \ge N_1<br />

Now, if m &gt; N_1,

<br /> \left| \frac{a_1 + a_2 + \dots + a_{N_1} + a_{N_1+1} + \dots + a_m}{m} - c \right| \le \left|\frac{(a_1-c) + (a_2 - c) + \dots + (a_{N_1} - c)}{m}\right| + \left|\frac{(a_{N_1+1} - c) + \dots + (a_m - c)}{m} \right|<br />

You can use the triangle inequality (repeatedly) on each of these two terms. Every piece from the second fraction can be bounded in relation to \epsilon, and since there are finitely many terms from the first piece, you can bound those. With a little though you should be able to show that, for m big enough, \sigma_m (the general term in the sum sequence) is within \epsilon of c

 

[JG89]

Ok. So let e denote epsilon.

I have

|(a_1 + .. + a_m)/m - c| <= |(a_1 - c)/m| + ... + |(a_m - c)/m| < |(a_1 - c)/m| + ... +

|(a_N1 - c)/m) + |m*e/m| = |(a_1 - c)/m| + ... + |(a_N1 - c)/m) + e. Now, assuming a_N1 is the largest term in the sequence a_1,...,a_N1, then, |(a_1 - c)/m| + ... + |(a_N1 - c)/m) + e
< |m(a_N1 - c)/m| + e = |a_N1 - c| + e. I know that for any n >= N1, |a_n - c| < e. So I know that |a_N1 - c| < e, so therefore |a_N1 - c| + e < e + e = 2e.

I'm stuck on that part now.