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Bounded Variation Proof

  1. Jul 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A sequence b_n is said to be of bounded variation if the series [tex] \sum_{n=1}^{\infty} |b_{n+1} - b_n| [/tex] converges.

    Prove that if b_n is of bounded variation, then the sequence b_n converges.


    2. Relevant equations



    3. The attempt at a solution

    If b_n is of bounded variation, then for all epsilon > 0, [tex] \sum_{v=n}^m |b_{n+1} - b_n| = |b_{n+1} - b_n| + |b_{n+2} - b_{n+1}| + ... + |b_m - b_{m-1}| + |b_{m+1} - b_m| < \epsilon [/tex] provided that n and m are sufficiently large.

    Notice that by the triangle inequality [tex] |b_{n+1} - b_n + b_{n+2} - b_{n+1} + ... + b_m - b_{m-1} + b_{m+1} - b_m| = |-b_n + b_{m+1}| = |b_n - b_{m+1}| \le |b_{n+1} - b_n| + |b_{n+2} - b_{n+1}| + ... + |b_m - b_{m-1}| + |b_{m+1} - b_m| < \epsilon [/tex] and so the sequence b_n is Cauchy, meaning it must converge. QED
     
  2. jcsd
  3. Jul 9, 2009 #2
    I have another question regarding bounded variations that is kind of similar to this question. The proof is short and I don't want to clog up the forum with multiple threads on the same topic, so I will post is in here:


    Question: If a sequence b_n is bounded and monotonic, prove that it is of bounded variation.

    Proof:

    In this proof we will assume the sequence is monotonic increasing -- the proof can easily be adapted to prove the same result for the sequence being monotonic decreasing.


    If b_n is bounded then there exists a fixed value, say B, such that |b_n| <= B for all n. Now,
    [tex] \sum_{v=1}^n |b_{n+1} - b_n| = |b_2 - b_1| + |b_3 - b_2| + |b_4 - b_3| + ... + |b_{n+1} - b_n| [/tex] . Now since b_n is increasing, b_{n+1} - b_n >= 0, and so we can remove all of the absolute value signs. We now have [tex] \sum_{v=1}^n |b_{n+1} - b_n| = |b_2 - b_1| + |b_3 - b_2| + |b_4 - b_3| + ... + |b_{n+1} - b_n| = b_2 - b_1 + b_3 - b_2 + ... + b_{n+1} - b_n = -b_1 + b_{n+1} \le b_1 + b_{n+1} \le 2B [/tex].

    Notice that the bound we've obtained for the n'th partial sum is a fixed value independent of n, and so [tex] 0 \le \sum_{v=1}^{\infty} |b_{n+1} - b_n| \le 2B [/tex]. The n'th partial sum is bounded and since each term is positive, the sequence of partial sums is increasing. Thus the sum converges.
     
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