The Squeeze Theorem

A sequence trapped between two others takes the stage. When both sides head to the same value, the middle follows.

The middle sequence never needs measuring directly. If both guards walk through the same door, whatever stands between them goes through it as well. Oscillating sequences are caught this way too: ∣sin⁡n∣≤1|\sin n| \le 1 traps sin⁡nn\frac{\sin n}{n} between ±1n\pm\frac{1}{n}, and since both sides tend to 0, so does the middle.

For example, picture a 🧍 person walking down a corridor. If both walls narrow in to a single point, the person, unable to pass through walls, must end up at that same point.

an≤bn≤cn∀na_n \le b_n \le c_n \quad \forall n
an→L, cn→L ⟹ bn→La_n \to L, \ c_n \to L \ \Longrightarrow \ b_n \to L

Examples & Exercises

Prove the squeeze theorem completely: if an≤bn≤cna_n \le b_n \le c_n for every nn and an→La_n \to L, cn→Lc_n \to L, then bn→Lb_n \to L.

an≤bn≤cn,an→L, cn→L ⟹ bn→La_n \le b_n \le c_n, \quad a_n \to L, \ c_n \to L \ \Longrightarrow \ b_n \to L
sin⁡nn\frac{\sin n}{n} — both guards close in on 0.
Show proof

  1. Let ε>0\varepsilon > 0 be given. Since an→La_n \to L, there is N1N_1 with ∣an−L∣<ε|a_n - L| < \varepsilon for n>N1n > N_1, i.e. L−ε<anL - \varepsilon < a_n. Likewise cn→Lc_n \to L gives N2N_2 with cn<L+εc_n < L + \varepsilon for n>N2n > N_2.

  2. Put N=max⁡(N1,N2)N = \max(N_1, N_2). For n>Nn > N both conditions hold at once, and combining with the trapping inequality gives L−ε<an≤bn≤cn<L+εL - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon. Hence ∣bn−L∣<ε|b_n - L| < \varepsilon.

  3. Since ε\varepsilon was arbitrary, bn→Lb_n \to L. The proof is complete.

Using the squeeze theorem, prove completely that lim⁡n→∞sin⁡nn=0\lim_{n \to \infty} \frac{\sin n}{n} = 0.

−1n≤sin⁡nn≤1n-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}
More terms: still oscillating, but closing in on 0.
Show proof

  1. For every real tt, ∣sin⁡t∣≤1|\sin t| \le 1. Dividing by n≥1n \ge 1 gives −1n≤sin⁡nn≤1n-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}.

  2. Both ends tend to 0: 1n→0\frac{1}{n} \to 0 and −1n→0-\frac{1}{n} \to 0 (the Archimedean corollary). The squeeze theorem now gives sin⁡nn→0\frac{\sin n}{n} \to 0.

Using the squeeze theorem, prove completely that lim⁡n→∞n+(−1)nn=1\lim_{n \to \infty} \frac{n + (-1)^n}{n} = 1.

n−1n≤n+(−1)nn≤n+1n\frac{n - 1}{n} \le \frac{n + (-1)^n}{n} \le \frac{n + 1}{n}
Even with the oscillating term, the ratio is squeezed to 1.
Show proof

  1. Since (−1)n(-1)^n is either −1-1 or 11, −1≤(−1)n≤1-1 \le (-1)^n \le 1. Adding nn and dividing by nn gives n−1n≤n+(−1)nn≤n+1n\frac{n - 1}{n} \le \frac{n + (-1)^n}{n} \le \frac{n + 1}{n}.

  2. The left end is n−1n=1−1n→1\frac{n - 1}{n} = 1 - \frac{1}{n} \to 1 and the right end is n+1n=1+1n→1\frac{n + 1}{n} = 1 + \frac{1}{n} \to 1. By the squeeze theorem the middle tends to 1 as well.

3. Sequences