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Limit of arithmetic mean

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

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

    3. 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!
  2. jcsd
  3. Oct 28, 2008 #2


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    Homework Helper

    Try this. Select [tex] \epsilon > 0 [/tex]. Since the original sequence converges to [tex] c [/tex] there is an integer [tex] N_1 [/tex] such that

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

    Now, if [tex] m > N_1 [/tex],

    \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 [tex] \epsilon [/tex], 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 [tex] m [/tex] big enough, [tex] \sigma_m [/tex] (the general term in the sum sequence) is within [tex] \epsilon [/tex] of [tex] c [/tex]
  4. Oct 28, 2008 #3
    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.
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