# 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:

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