Comparison Theorems for Sequences

# Comparison Theorems for Sequences

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

 Theorem 1: If $a_n ≤ b_n$ for all $n ≥ N$ 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 $n ≥ N$ (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 $n ≥ N$ then $a_n > k$.
• But if $a_n ≤ b_n$ for all $n ≥ N$, then we get that $k < a_n ≤ b_n$ if $n ≥ N$ or rather $k < b_n$. So by the definition, $\lim_{n \to \infty} b_n = \infty$. $\blacksquare$
 Theorem 2: If $a_n ≥ b_n$ for all $n ≥ N$ 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 $n ≥ N$, 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 $a_n ≥ b_n$ or rather $b_n ≤ a_n < k$ for all $n ≥ N$ , then we get that $b_n < k$ if $n ≥ N$. 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 $a_n ≤ b_n$ for all $n ≥ N$ then $A ≤ B$.

What this theorem says is that if the sequence $\{ b_n \}$ is ultimately above the sequence $\{ a_n \}$ when $n ≥ N$, 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 $n ≥ N_1$ 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 $n ≥ N_2$ then $\mid b_n - B \mid < \frac{A - B}{3} = \epsilon$.
• Now we want to pick an $N \in \mathbb{N}$ such that $N ≥ N_1$ and $N ≥ N_2$ 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 $n ≥ N$, then $a_n ≤ b_n$ 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 $A ≤ B$. $\blacksquare$