Squeeze Theorem for Sequences

Welcome to our next lesson on sequences. In our previous session, we built a powerful toolkit using algebraic limit laws and L'Hôpital's rule to find the limits of many sequences. However, we ended with a puzzle: how do we handle sequences with oscillating terms, like those involving $sin (n)$ or $(- 1)^{n}$ , where our previous methods fall short?

This lesson introduces the elegant and intuitive tool designed for precisely these situations: the Squeeze Theorem for sequences. You will learn to state the theorem, understand its formal proof, and, most importantly, apply it to determine the limits of sequences that are "squeezed" between two more cooperative ones. This will round out your essential techniques for evaluating sequence limits, a crucial skill as we build towards understanding infinite series.

The Intuition Behind the Squeeze

The core idea of the Squeeze Theorem is as simple as its name suggests. Imagine you have a sequence whose terms are always trapped between the terms of two other sequences. If you can show that the two outer "bounding" sequences both converge to the exact same limit, then the sequence in the middle has no choice but to be squeezed toward that very same limit.

The following image provides a perfect visual for this concept.

The Formal Statement

While the intuition is straightforward, mathematics demands precision. Let's look at the formal statement of the theorem.

5.1 Sequences - Calculus Volume 2

Please read the formal statement of the Squeeze Theorem and its accompanying proof.

In the text, locate the section for Theorem 5.4, Squeeze Theorem for Sequences. Read the theorem statement carefully. Note the three crucial conditions: the inequality that must hold and the fact that both bounding sequences must converge to the same limit, L. You can also read the associated proof, which uses the formal \epsilon-N definition of a limit.

As you read, the key takeaway is that to use the theorem, you must satisfy all its conditions. You need to:

Find a lower-bounding sequence, $a_{n}$ .
Find an upper-bounding sequence, $c_{n}$ .
Show that $a_{n} \leq b_{n} \leq c_{n}$ for all $n$ after some point.
Show that $lim_{n \to \infty} a_{n} = L$ and $lim_{n \to \infty} c_{n} = L$ for some limit $L$ .

Only if all four conditions are met can you conclude that $lim_{n \to \infty} b_{n} = L$ .

Applying the Squeeze Theorem: A Worked Example

Let's see how this works in practice by tackling one of the problems we couldn't solve in the last lesson: finding the limit of $a_{n} = \frac{s i n ( n )}{n ^{2}}$ .

The following video provides an excellent walkthrough of this exact type of problem, showing how to establish the bounds and apply the theorem.

Squeeze Theorem For Sequences

This video from The Organic Chemistry Tutor demonstrates how to apply the Squeeze Theorem to a sequence with a trigonometric term.

Watch the segment from this example. The presenter clearly explains how the known bounds of the sine function are used to create the "squeezing" sequences.

The strategy demonstrated here is fundamental. The most challenging part of using the Squeeze Theorem is often finding the bounding sequences. For sequences involving trigonometric functions like sine or cosine, the key is to use their natural bounds:

$- 1 \leq sin (x) \leq 1$
$- 1 \leq cos (x) \leq 1$
$0 \leq sin^{2} (x) \leq 1$
$0 \leq cos^{2} (x) \leq 1$

By starting with these simple inequalities, you can build up the more complex expression of your sequence and derive your two bounding sequences.

The Rigorous Proof

The proof in the OpenStax textbook you read earlier is concise and correct. However, given your interest in rigorous understanding, I recommend watching a more detailed, step-by-step construction of the proof. This video does an exceptional job of walking through the formal $ϵ$ - $N$ logic, showing exactly how the convergence of the outer sequences forces the convergence of the inner one.

Proof: Sequence Squeeze Theorem | Real Analysis

This video from Wrath of Math provides a detailed, formal proof of the Squeeze Theorem for sequences.

First, watch the brief statement of the theorem. Then, skip to the main proof, which begins at this timestamp. Pay close attention to how the definition of convergence is applied to the bounding sequences a_n and b_n, and how the choice of N = \max(N_1, N_2) is used to guarantee that all three sequences are behaving as needed.

This proof beautifully connects the intuitive idea of "squeezing" to the formal definition of a limit. It demonstrates that for any desired closeness $ϵ$ , we can find a point $N$ in the sequences after which the middle sequence $x_{n}$ is guaranteed to be within $ϵ$ of the limit $L$ , because the sequences holding it in place are.

Try It Yourself

Now it's your turn to apply the Squeeze Theorem. For each sequence below, find two bounding sequences and use the Squeeze Theorem to determine its limit.

$a_{n} = \frac{( - 1 ) ^{n}}{n ^{2}}$
$b_{n} = \frac{2 n + c o s ( n )}{3 n - 1}$

Hint for $b_{n}$ : First, find the bounds for the numerator, $2 n + cos (n)$ . Then divide the entire inequality by the denominator, $3 n - 1$ . You will then need to find the limits of the new bounding sequences.

Solutions:

Sequence $a_{n} = \frac{( - 1 ) ^{n}}{n ^{2}}$ :
The term $(- 1)^{n}$ oscillates between -1 and 1. This gives us our starting inequality.
$- 1 \leq (- 1)^{n} \leq 1$
Since $n^{2}$ is always positive for $n \geq 1$ , we can divide the entire inequality by $n^{2}$ without changing the direction of the inequalities:
$- \frac{1}{n ^{2}} \leq \frac{( - 1 ) ^{n}}{n ^{2}} \leq \frac{1}{n ^{2}}$
We have found our bounding sequences: the lower bound is $- \frac{1}{n ^{2}}$ and the upper bound is $\frac{1}{n ^{2}}$ . Now we check their limits:
$lim_{n \to \infty} (- \frac{1}{n ^{2}}) = 0$
$lim_{n \to \infty} (\frac{1}{n ^{2}}) = 0$
Since the sequence $a_{n}$ is squeezed between two sequences that both converge to 0, we conclude by the Squeeze Theorem that $lim_{n \to \infty} a_{n} = 0$ .
Sequence $b_{n} = \frac{2 n + c o s ( n )}{3 n - 1}$ :
We start with the bounds for the $cos (n)$ term:
$- 1 \leq cos (n) \leq 1$
Now, we build up the numerator by adding $2 n$ to all parts:
$2 n - 1 \leq 2 n + cos (n) \leq 2 n + 1$
Next, we divide by the denominator, $3 n - 1$ . Since $n \geq 1$ , the denominator is positive, so the inequalities are preserved:
$\frac{2 n - 1}{3 n - 1} \leq \frac{2 n + c o s ( n )}{3 n - 1} \leq \frac{2 n + 1}{3 n - 1}$
Now we find the limits of our bounding sequences using the techniques from the last lesson (dividing by the highest power of $n$ ):
$lim_{n \to \infty} \frac{2 n - 1}{3 n - 1} = lim_{n \to \infty} \frac{2 - 1/ n}{3 - 1/ n} = \frac{2}{3}$
$lim_{n \to \infty} \frac{2 n + 1}{3 n - 1} = lim_{n \to \infty} \frac{2 + 1/ n}{3 - 1/ n} = \frac{2}{3}$
Both the lower and upper bounds converge to $\frac{2}{3}$ . Therefore, by the Squeeze Theorem, $lim_{n \to \infty} b_{n} = \frac{2}{3}$ .

Conclusion

In this lesson, you've added a vital and versatile tool to your mathematical arsenal. The Squeeze Theorem provides a rigorous way to determine the limit of a sequence by trapping it between two simpler, convergent sequences.

The key takeaways are:

The Squeeze Theorem is ideal for sequences with oscillating or bounded terms (like $sin (n)$ , $cos (n)$ , or $(- 1)^{n}$ ) that are otherwise difficult to evaluate.
The strategy involves finding upper and lower bounding sequences whose limits are known and, crucially, identical.
The proof relies on the formal $ϵ$ - $N$ definition of a limit to show that the middle sequence is forced to converge.

So far, we have established convergence by either directly calculating the limit (using algebra or L'Hôpital's rule) or by "forcing" it with the Squeeze Theorem. But what if a sequence converges, yet we can't easily calculate its limit? In our next lesson, we will explore a different way to prove convergence by looking at the internal properties of a sequence itself—specifically, whether it is monotonic and bounded.