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 $$\epsilon > 0$$. Since the original sequence converges to $$c$$ there is an integer $$N_1$$ such that

$$|a_n - c | < \epsilon \quad \text{ for } n \ge N_1$$

Now, if $$m > N_1$$,

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

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.