The Timpson Lyceum · Real Analysis (2020)

Lecture 7 (Real Analysis 2020) Part 2

Overview

This lecture develops the Monotone Convergence Theorem, one of the central results of sequence theory: a monotone sequence converges exactly when it is bounded. We first define what it means for a sequence to be monotonically increasing, monotonically decreasing, or simply monotone, then prove the increasing case of the theorem in full. With the theorem in hand, we apply it three times: to show that a bounded increasing series converges, to show that the harmonic series diverges, and to prove that $c^n \to 0$ whenever $|c| < 1$ . The power of the result is that it certifies convergence without our ever needing to know the actual limit.

Key Concepts

Monotone sequences. A sequence $(a_n)$ is:

Monotonically increasing if $a_{n+1} \ge a_n$ for all $n$ (each term stays the same or grows).
Monotonically decreasing if $a_{n+1} \le a_n$ for all $n$ (each term stays the same or shrinks).
Monotone if it is one of the two. (A sequence that is both increasing and decreasing is exactly a constant sequence.)

Standard test for monotonicity. To classify a sequence, examine the difference $a_{n+1} - a_n$ :

$a_{n+1} - a_n \ge 0$ for all $n$ $\;\Rightarrow\;$ increasing,
$a_{n+1} - a_n \le 0$ for all $n$ $\;\Rightarrow\;$ decreasing.

Monotone Convergence Theorem. Let $(a_n)$ be monotonically increasing. Then $\lim_{n \to \infty} a_n = \sup_n a_n \quad\Longleftrightarrow\quad (a_n)\text{ is bounded above}.$ Equivalently, a monotonically increasing sequence converges if and only if it is bounded above, and its limit is then its supremum. The decreasing case is identical: a monotonically decreasing sequence converges iff it is bounded below, and its limit is its infimum.

Easy direction ( $\Rightarrow$ ): If $\lim a_n = \sup a_n$ , the sequence converges, and every convergent sequence is bounded (proved earlier).
Hard direction ( $\Leftarrow$ ): Boundedness $+$ monotonicity $\Rightarrow$ convergence to the supremum. This is where the Completeness Axiom enters — a set bounded above has a least upper bound.

Why the supremum is the limit. Because $\sup$ is the least upper bound, $B - \varepsilon$ (for any $\varepsilon > 0$ ) is not an upper bound, so some term $a_N$ exceeds it. Monotonicity then drags every later term above $B - \varepsilon$ too, trapping the tail inside $(B - \varepsilon, B]$ .

Arithmetic of sequences. Sequences add and multiply termwise, exactly as expected: $(a_n) + (b_n)$ has $n$ th term $a_n + b_n$ , and $(a_n)(b_n)$ has $n$ th term $a_n b_n$ . A series is a sequence built by summing terms of another sequence, $S_n = \sum_{k=1}^{n} c_k$ .

Geometric sum formula (used in the applications): $\sum_{k=0}^{n-1} r^k = \frac{1 - r^n}{1 - r}.$

Worked Example

Proof of the hard direction. Let $(a_n)$ be monotonically increasing and bounded above. Let $B = \sup_n a_n$ , which exists by completeness. Fix $\varepsilon > 0$ .

\begin{aligned} B - \varepsilon &< B = \sup_n a_n &&\Rightarrow\; B-\varepsilon\text{ is not an upper bound} \\ &\exists\, N:\; a_N > B - \varepsilon \\ n \ge N &\;\Rightarrow\; a_n \ge a_N > B - \varepsilon &&\text{(monotone increasing)} \\ B - \varepsilon &< a_n \le B < B + \varepsilon &&(B\text{ is an upper bound)} \\ \Rightarrow\quad |a_n - B| &< \varepsilon \quad \text{for all } n \ge N. \end{aligned}

Since $\varepsilon > 0$ was arbitrary, $\lim_{n\to\infty} a_n = B = \sup_n a_n$ . $\blacksquare$

Application — a bounded series converges. Let $S_n = \sum_{k=1}^{n} \frac{1}{k}\cdot\frac{1}{2^k}$ . Each added term is positive, so $S_{n+1} - S_n = \frac{1}{(n+1)2^{n+1}} > 0$ , making $(S_n)$ increasing. Bounding above by dropping the $\frac1k$ factor and summing a geometric series:

\begin{aligned} S_n &= \sum_{k=1}^{n}\frac{1}{k}\cdot\frac{1}{2^k} \;<\; \sum_{k=1}^{n}\frac{1}{2^k} \\ &= \frac{\tfrac12\left(1 - \left(\tfrac12\right)^n\right)}{1 - \tfrac12} = 1 - \left(\tfrac12\right)^n \;\le\; 1. \end{aligned}

So $(S_n)$ is increasing and bounded above by $1$ ; by the theorem it converges. $\blacksquare$

Summary

A sequence is monotone if it is monotonically increasing ( $a_{n+1}\ge a_n$ ) or monotonically decreasing ( $a_{n+1}\le a_n$ ); a constant sequence is both. Classify by the sign of $a_{n+1}-a_n$ .
Monotone Convergence Theorem: an increasing sequence converges $\iff$ it is bounded above, with limit $\sup a_n$ ; a decreasing sequence converges $\iff$ it is bounded below, with limit $\inf a_n$ . The hard direction relies on the Completeness Axiom.
The theorem certifies convergence without revealing the limit — useful when the limit is hard to compute.
A bounded, increasing series such as $\sum \frac{1}{k\,2^k}$ converges because it stays below $1$ .
The harmonic series $\sum_{k=1}^{n}\frac{1}{k}$ diverges: grouping terms shows $S_{2^k} \ge 1 + \tfrac{k}{2}$ , which is unbounded, so the series cannot converge.
For $|c| < 1$ , $\lim_{n\to\infty} c^n = 0$ : for $0<c<1$ the sequence $c^n$ is decreasing and bounded below by $0$ , so it converges to its infimum, which a contradiction argument shows must be $0$ ; the cases $c=0$ and $c<0$ follow by reduction to $|c|^n$ .