Squeeze Theorem for Sequences¶

Theorem¶

Let $(a_n)_{n=1}^\infty\text{,}$ $(b_n)_{n=1}^\infty\text{,}$ and $(c_n)_{n=1}^\infty$ be sequences of real numbers such that $a_n\leq b_n\leq c_n$ for all $n \geq 1\text{.}$ Suppose also that $(a_n)_{n=1}^\infty$ and $(c_n)_{n=1}^\infty$ converge to the same limit $L\text{.}$ Then $(b_n)_{n=1}^\infty$ also converges to $L\text{.}$

Interpretation¶

The squeeze theorem proves convergence by trapping: a sequence caught between two sequences with the same limit has nowhere else to go. It is the standard tool when $b_n$ is hard to handle directly but easy to bound by simpler sequences.

Three sequences approaching a limit L: an upper one from above, a lower one from below, and an oscillating middle one trapped between them.

In the figure, $c_n$ approaches $L$ from above and $a_n$ from below, while the middle sequence $b_n$ oscillates around $L$ . The oscillation does not matter: $b_n$ always stays between $a_n$ and $c_n$ , and the gap between them shrinks to $0$ .

Shrink $\varepsilon$ : from the index where both $a_n$ and $c_n$ are inside the band, $b_n$ is trapped inside too, so it converges to the same $L$ .

Both hypotheses are needed. The outer limits must be equal: $b_n=1+(-1)^n$ lies between $a_n=0$ and $c_n=2$ but alternates between $0$ and $2$ . Both bounds are needed too: $b_n=-n$ satisfies $b_n\le\frac{1}{n}$ with $\frac{1}{n}\to0$ $0$ , yet diverges. The inequalities only have to hold from some index on, since finitely many terms do not affect a limit. Unlike the monotone convergence theorem, the theorem needs no monotonicity, and it gives both the existence and the value of the limit.

In practice the theorem is often used in the form $|b_n-L|\le d_n$ with $d_n\to0$ $0$ , which is the theorem with $a_n=L-d_n$ and $c_n=L+d_n$ $d_{n}$ . For example, $\left|\frac{(-1)^n}{n}\right|\le\frac{1}{n}$ gives $\frac{(-1)^n}{n}\to0$ $0$ . The same trapping argument returns later for functions in the squeeze theorem for functions.

Key Results¶

The middle sequence $(b_n)_{n=1}^\infty$ must lie termwise between two convergent sequences.
Equality of outer limits enforces convergence of the middle sequence.
The theorem does not require monotonicity.

Examples¶

If $-1/n \le b_n \le 1/n\text{,}$ then $b_n\to 0\text{.}$
$\lim\limits_{n\to\infty}\left(n \cdot\sin(1/n)\right)= 1\text{.}$

Prerequisites¶

Sources¶

Math I: Calculus · Lecture 3

Aliases: sandwich theorem for sequences Keywords: sandwich theorem, squeeze theorem Symbols: epsilon, infinity

Last updated: 20 October 2026 03:02