Comparison Theorems for Sequences

Table of Contents

We will now look at a few very important comparison theorems for both divergent and convergent sequences.

Theorem 1: If

$a_n \leq b_n$

for all

$n \geq N$

and

\lim_{n \to \infty} a_n = \infty

then

\lim_{n \to \infty} b_n = \infty

What theorem 1 essentially says is that if a sequence $\{ a_n \}$ races off to infinity, and the sequence $\{ b_n \}$ is above the sequence $\{ a_n \}$ when $$n \geq N$$ (that is $\{ b_n \}$ is ultimately above $\{ a_n \}$ ), then the sequence $\{ b_n \}$ must also race off to infinity.

Proof of Theorem 1: We know that $\lim_{n \to \infty} a_n = \infty$ implies that $\forall k \in \mathbb{R} \: \exists N \in \mathbb{N}$ such that if $$n \geq N$$ then $$a_n > k$$ .

But if $$a_n \leq b_n$$ for all $$n \geq N$$ , then we get that $$k < a_n \leq b_n$$ if $$n \geq N$$ or rather $$k < b_n$$ . So by the definition, $\lim_{n \to \infty} b_n = \infty$ . $\blacksquare$

Theorem 2: If

$a_n \geq b_n$

for all

$n \geq N$

and

\lim_{n \to \infty} a_n = -\infty

then

\lim_{n \to \infty} b_n = -\infty

This theorem is very similar to theorem 1. It says that if the sequence $\{ a_n \}$ races off to negative infinity, and the seqeunce $\{ b_n \}$ is below the sequence $\{ a_n \}$ when $$n \geq N$$ , then the sequence $\{ b_n \}$ must also race off to negative infinity.

Proof of Theorem 2: We know that $\lim_{n \to \infty} a_n = -\infty$ implies that if $\forall k \in \mathbb{R} \: \exists N \in \mathbb{N} $}] such that if [[$ n ≥ N$ then $$a_n < k$$ .

But if $$a_n \geq b_n$$ or rather $$b_n \leq a_n < k$$ for all $$n \geq N$$ , then we get that $$b_n < k$$ if $$n \geq N$$ . So by the definition, $\lim_{n \to \infty} b_n = -\infty$ . $\blacksquare$

Theorem 3: If

\lim_{n \to \infty} a_n = A

and

\lim_{n \to \infty} b_n = B

and

$a_n \leq b_n$

for all

$n \geq N$

then

$A \leq B$

What this theorem says is that if the sequence $\{ b_n \}$ is ultimately above the sequence $\{ a_n \}$ when $$n \geq N$$ , and both sequences are convergent, then the point to which $\{ b_n \}$ converges must be above or equal to the point to which $\{ a_n \}$ converges.

Proof of Theorem 3: Suppose that in fact $$A > B$$ . We want to establish a contradiction here so we will let $\epsilon = \frac{A - B}{3}$ .

So $\lim_{n \to \infty} a_n$ implies that $\forall \epsilon > 0 \: \exists N_1 \in \mathbb{N}$ such that if $$n \geq N_1$$ then $\mid a_n - A \mid < \frac{A - B}{3} = \epsilon$ .

Similarly $\lim_{n \to \infty} b_n$ implies that $\forall \epsilon > 0 \: \exists N_2 \in \mathbb{N}$ such that if $$n \geq N_2$$ then $\mid b_n - B \mid < \frac{A - B}{3} = \epsilon$ .

Now we want to pick an $N \in \mathbb{N}$ such that $$N \geq N_1$$ and $$N \geq N_2$$ and thus we choose $N = \mathrm{max} \{ N_1, N_2 \}$ .

So we know that $\mid a_n - A \mid < \frac{A - B}{3}$ implies that $- \frac{A-B}{3} < a_n - A < \frac{A-B}{3}$ . So then $A - \frac{A - B}{3} < a_n$ .

Similarly we know that $\mid b_n - B \mid < \frac{A - B}{3}$ implies that $- \frac{A-B}{3} < b_n - B < \frac{A-B}{3}$ . So then $b_n < B + \frac{A-B}{3}$ .

So we know that if $$n \geq N$$ , then $$a_n \leq b_n$$ and so:

(1)

\begin{align} \quad A - \frac{A - B}{3} < a_n ≤ b_n < B + \frac{A - B}{3} \\ A - \frac{A - B}{3} ≤ B + \frac{A - B}{3} \\ 3A - (A - B) ≤ 3B + (A - B) \\ 2A + B ≤ A + 2B \\ A + B ≤ 2B \\ A ≤ B \end{align}

But that's a contradiction since we assumed that $$A > B$$ . Therefore $$A \leq B$$ . $\blacksquare$

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Comparison Theorems for Sequences