The Squeeze Theorem for Convergent Sequences

We will now proceed to specifically look at the limit squeeze theorem (law 7 from the Limit of a Sequence page) and prove it's validity.

Law 7 (Squeeze Theorem for Convergent Sequences): If the limits of the sequences $\{ a_n \}$, $\{ b_n \}$, and $\{ c_n \}$ are convergent and $a_n ≤ b_n ≤ c_n$ is true always after some $n^{\mathrm{th}}$ term, if $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$, then $\lim_{n \to \infty} b_n = L$.

• Proof of Law 7: Let $\{ a_n \}$, $\{ b_n \}$ and $\{ c_n \}$ be convergent sequences such that $a_n ≤ b_n ≤ c_n$ is true always after some $M^{\mathrm{th}}$ term, and let $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$. We want to show that $\forall \epsilon > 0 \: \exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid b_n - L \mid < \epsilon$.

• We note that $\lim_{n \to \infty} a_n = L$ implies that $\forall \epsilon > 0 \: \exists N_1$ such that if $n ≥ N_1$ then $\mid a_n - L \mid < \epsilon$ or rather, $-\epsilon < a_n - L < \epsilon$.

• Similarly we note that $\lim_{n \to \infty} c_n = L$ implies that $\forall \epsilon > 0 \: \exists N_2$ such that if $n ≥ N_2$ then $\mid c_n - L \mid < \epsilon$ or rather $-\epsilon < c_n - L < \epsilon$.

• Now let $N = \mathrm{max} \{ M, N_1, N_2 \}$ to ensure that $-\epsilon < a_n - L < \epsilon$, $-\epsilon < c_n - L \epsilon$, and $a_n ≤ b_n ≤ c_n$. Subtracting $L$ from all parts of this inequality we get that $a_n - L ≤ b_n - L ≤ c_n - L$ or rather $-\epsilon < b_n - L < \epsilon$ so then $\mid b_n - L \mid < \epsilon$ and thus $\lim_{n \to \infty} b_n = L$. $\blacksquare$

The Squeeze Theorem is an important result because we can determine a sequence's limit if we know it is "squeezed" between two other sequences whose limit is the same. We will now look at another important theorem proven from the Squeeze Theorem.

Theorem 1: If $\lim_{n \to \infty} \mid a_n \mid = 0$ then $\lim_{n \to \infty} a_n = 0$.

• Proof of Theorem 1: We first note that $-\mid a_n \mid ≤ a_n ≤ \mid a_n \mid$. We are given that $\lim_{n \to \infty} \mid a_n \mid = 0$ and similarly $\lim_{n \to \infty} -\mid a_n \mid = 0$. By the squeeze theorem, it follows that $\lim_{n \to \infty} a_n = 0$.

Example 1

Determine the limit of the sequence $\left \{ \frac{\cos n}{n} \right \}$.

First note that $-1 ≤ \cos n ≤ 1$ for all $n \in \mathbb{N}$. Therefore we have that:

(1)

\begin{align} \quad -\frac{1}{n} ≤ \frac{\cos n}{n} ≤ \frac{1}{n} \end{align}

Now $\lim_{n \to \infty} \frac{1}{n} = 0$ and $\lim_{n \to \infty} -\frac{1}{n} = 0$, and so by the Squeeze Theorem we have that $\lim_{n \to \infty} \frac{\cos n}{n} = 0$.