Monotone Convergence Theorem
A monotone sequence is one that never changes direction (always increasing or always decreasing). Theorem: A monotone sequence converges if and only if it is bounded.
Drag the Red Line (Upper Bound)
Intuition: If you are walking up a staircase that never goes down, and there is a ceiling you cannot pass, you must eventually get infinitely close to some height (the supremum).
Cauchy Sequences
A sequence is Cauchy if its terms eventually get arbitrarily close to each other. This definition doesn't require knowing the limit \( L \) in advance!
\( (s_n) \) is Cauchy if \( \forall \epsilon \gt 0, \exists N \) such that \( \forall n, m \gt N, |s_n - s_m| \lt \epsilon \).
Cauchy Inspector
Adjust \( \epsilon \) to see the "tail" of the sequence where all terms are clustered together.
For all \( n, m \gt N \), the distance between points is less than \( \epsilon \).
Practice Problems
True/False
Every bounded sequence is convergent.
False.
Counterexample: \( s_n = (-1)^n \) is bounded by 1 but diverges (oscillates between -1 and 1).
Every Cauchy sequence is bounded.
True.
This is Lemma 10.10. Cauchy sequences eventually cluster, so the tail is bounded, and the finite number of initial terms are also bounded.
If a sequence is monotone, it must converge.
False.
It must be bounded as well. Counterexample: \( s_n = n \) is monotone (increasing) but diverges to \( +\infty \).
If \( \lim s_n = s \), then \( (s_n) \) is a Cauchy sequence.
True.
Convergent sequences are always Cauchy (in any metric space). The terms get close to the limit, so they must get close to each other.
Fill in the Blank
A sequence \( (s_n) \) is increasing if \( s_n \le s_{n+1} \) for all \( n \in \mathbb{N} \).
Answer: \( s_n \le s_{n+1} \)
The Monotone Convergence Theorem states that every ________ monotone sequence converges.
Answer: bounded
A sequence \( (s_n) \) is Cauchy if for every \( \epsilon \gt 0 \), there exists \( N \) such that for all \( n, m \gt N \), ________.
Answer: \( |s_n - s_m| \lt \epsilon \)
If a sequence is unbounded and increasing, its limit is ________.
Answer: \( +\infty \)
Full Problems
Prove that the sequence defined by \( s_1 = 1 \) and \( s_{n+1} = \sqrt{2s_n} \) is convergent.
Strategy: Use the Monotone Convergence Theorem. Show it is bounded above by 2 and increasing.
- Bounded: Base case \( s_1 = 1 \lt 2 \). Assume \( s_k \lt 2 \). Then \( s_{k+1} = \sqrt{2s_k} \lt \sqrt{2(2)} = 2 \). By induction, bounded above by 2.
- Increasing: \( s_{n+1} \ge s_n \iff \sqrt{2s_n} \ge s_n \iff 2s_n \ge s_n^2 \iff 2 \ge s_n \) (since \( s_n \gt 0 \)). Since we proved \( s_n \lt 2 \), it is increasing.
- By MCT, it converges. (Limit is actually 2).
Let \( s_n = \frac{1}{n} \). Prove directly using the definition that \( (s_n) \) is a Cauchy sequence.
Let \( \epsilon \gt 0 \). Choose \( N \gt \frac{2}{\epsilon} \).
For any \( n, m \gt N \):
\( |s_n - s_m| = |\frac{1}{n} - \frac{1}{m}| \le \frac{1}{n} + \frac{1}{m} \lt \frac{1}{N} + \frac{1}{N} = \frac{2}{N} \lt \epsilon \).
Prove that if \( (s_n) \) is a Cauchy sequence, then \( (s_n) \) is bounded.
Let \( \epsilon = 1 \). Since Cauchy, \( \exists N \) such that \( n, m \gt N \implies |s_n - s_m| \lt 1 \).
Fix \( m = N+1 \). Then for all \( n \gt N \), \( |s_n - s_{N+1}| \lt 1 \implies |s_n| \lt |s_{N+1}| + 1 \).
Let \( M = \max\{|s_1|, |s_2|, \dots, |s_N|, |s_{N+1}| + 1\} \). Then \( |s_n| \le M \) for all \( n \).
Let \( s_n = \sum_{k=1}^n \frac{1}{k^2} \). Show that \( (s_n) \) converges.
Strategy: Show it is increasing and bounded.
- Increasing: \( s_{n+1} - s_n = \frac{1}{(n+1)^2} \gt 0 \).
- Bounded: \( \frac{1}{k^2} \lt \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k} \) for \( k \ge 2 \).
- Sum telescopes: \( s_n \lt 1 + \sum_{k=2}^n (\frac{1}{k-1} - \frac{1}{k}) = 1 + (1 - \frac{1}{n}) \lt 2 \).
- By MCT, it converges.