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Monotone Convergence Theorem and Infinite Series

MATH 4630 · Chapter 2: Sequences and Series

We now shift focus from general sequences to two important special classes: monotone sequences (which behave predictably) and infinite series (which are sums of infinitely many terms). The Monotone Convergence Theorem guarantees convergence under order structure, and series theory extends our convergence tools to new contexts.

The Monotone Convergence Theorem

Definition: Monotone Sequences

A sequence $(a_n)$ is:

Monotone increasing (or non-decreasing) if $a_n \leq a_{n+1}$ for all $n \in \mathbb{N}$;
Monotone decreasing (or non-increasing) if $a_n \geq a_{n+1}$ for all $n \in \mathbb{N}$;
Monotone if it is either monotone increasing or monotone decreasing.

Theorem: Monotone Convergence Theorem (MCT)

A monotone sequence $(a_n)$ converges if and only if it is .

Proof Sketch (for increasing bounded sequences)

Suppose $(a_n)$ is monotone increasing and bounded. Then the set $\{a_n : n \in \mathbb{N}\}$ is a bounded set of real numbers. By the Completeness Axiom, this set has a least upper bound (supremum).

Define the limit candidate:

Let $a = \sup\{a_n : n \in \mathbb{N}\}$.

Show $(a_n) \to a$:

Let $\varepsilon > 0$. Since $a - \varepsilon < a = \sup\{a_n\}$, there must exist some term $a_N$ such that

$$a - \varepsilon < a_N$$

Since $(a_n)$ is monotone increasing, for all $n \geq N$:

$$a - \varepsilon < a_N \leq a_n \leq a$$

Therefore:

Thus, $(a_n) \to a$. $\square$

Example: The Sequence $a_n = \dfrac{n}{n+1}$

Step 1: Show it's monotone increasing

We need to check if $a_n \leq a_{n+1}$:

$$\frac{n}{n+1} \leq \frac{n+1}{n+2}$$

Cross-multiply (both denominators are positive):

Step 2: Show it's bounded above

For all $n$: $a_n = \dfrac{n}{n+1} = \dfrac{n+1-1}{n+1} = 1 - \dfrac{1}{n+1} < 1$.

Step 3: Apply MCT

Since $(a_n)$ is monotone increasing and bounded above, by MCT it converges. The limit is

$$\lim_{n \to \infty} \frac{n}{n+1} = $$

Infinite Series

Definition: Series and Partial Sums

Let $(a_n)_{n=1}^{\infty}$ be a sequence. An infinite series is a formal sum

$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$$

The $N$-th partial sum is

$$S_N = \sum_{n=1}^{N} a_n = $$

The series converges to $S$ if the sequence of partial sums $(S_N)$ converges to $S$. We write

$$\sum_{n=1}^{\infty} a_n = S$$

Example: The Harmonic Series

The harmonic series is

$$\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$$

Question: Does this converge or diverge?

Consider the partial sum $S_{2^k}$ for $k = 1, 2, 3, \ldots$:

$$S_{2^k} = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots$$

Group terms strategically:

$$S_{2^k} \geq 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots$$

Simplify:

$$S_{2^k} \geq 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = 1 + \frac{k}{2}$$

As $k \to \infty$, we have $S_{2^k} \geq 1 + \dfrac{k}{2} \to \infty$. Therefore, the harmonic series .

Tests for Convergence of Series

Theorem: The Cauchy Condensation Test

Let $(a_n)$ be a non-negative decreasing sequence. Then

$$\sum_{n=1}^{\infty} a_n \quad \text{converges} \quad \Leftrightarrow \quad \sum_{k=0}^{\infty} 2^k a_{2^k} \quad \text{converges}$$

Theorem: The $p$-Series Test

For $p > 0$, the series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ converges if and only if $p > 1$.

Proof of $p$-Series Using Condensation Test

Let $p > 0$ and consider $a_n = \dfrac{1}{n^p}$. Then $a_n$ is positive and decreasing.

By the Condensation Test:

$$\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \quad \Leftrightarrow \quad \sum_{k=0}^{\infty} 2^k \cdot \frac{1}{(2^k)^p} \text{ converges}$$

Simplify the condensed series:

$$\sum_{k=0}^{\infty} 2^k \cdot 2^{-pk} = \sum_{k=0}^{\infty} 2^{k(1-p)}$$

This is a series with ratio $r = 2^{1-p}$.

A geometric series converges if and only if $|r| < 1$, so:

$$2^{1-p} < 1 \quad \Leftrightarrow \quad 1 - p < 0 \quad \Leftrightarrow \quad p > 1$$

Therefore, $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ converges if and only if $p > 1$. $\square$

Exercises

Exercise 1

For each of the following, determine if the series converges or diverges. Justify your answer.

$\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{0.5}}$
$\sum\limits_{n=1}^{\infty} \dfrac{1}{n^2}$
$\sum\limits_{n=1}^{\infty} \dfrac{1}{2n}$

Exercise 2

Consider the sequence $b_n = 1 - \dfrac{1}{2^n}$. Show that $(b_n)$ is monotone increasing and bounded, then find $\lim\limits_{n \to \infty} b_n$ using MCT.

Key Takeaway

Monotone Convergence Theorem: Order structure + boundedness ⟹ convergence (no need for $\varepsilon$-$N$ arguments).

Series: Extend convergence theory from sequences to infinite sums. Test convergence using specialized criteria like condensation or the $p$-series rule.

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