Does Monotonicity and Boundedness Imply Bounded Variation?

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Homework Statement



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.


Homework Equations





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
 
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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.