Squeeze theorem for sequences: Worked examples

The scenario matrix

Every squeeze problem falls into one of these cells. First, two words we use loosely in the table, made precise:

The examples that follow are labelled by cell so you can see the whole map is covered.

Cell	What makes it this cell	Wall strategy	Example
A. Bounded oscillator ÷ grower	a bounded oscillator ( $sin$ , $cos$ , $(- 1)^{n}$ ) divided by a grower ( $x_{n} \to \infty$ )	$\pm$ the bound over the grower	Ex 1
B. Sign-changing middle	the middle itself can be negative	must use a two-sided wall, never one	Ex 2
C. Product / factorial vs power	terms multiplied, one tames the rest	crude upper bound $\leq \frac{c}{n}$ , lower $0$	Ex 3
D. Root of a polynomial	$n$ th roots, $a^{1/ n}$	sandwich by known $n$ th roots ( $n n \to 1$ )	Ex 4
E. Sum of many small bits	a sum whose term count grows with $n$	bound every term by the same walls	Ex 5
F. Degenerate / equal walls	walls collapse onto the middle, or middle is a wall	direct read-off, watch for "walls to different limits"	Ex 6
G. Word / real-world	error term, physical bound	model the noise as bounded, squeeze	Ex 7
H. Exam twist	walls that look like they go to different limits but don't	simplify walls first, then squeeze	Ex 8

Prerequisites lean on: Limit of a sequence (epsilon-N definition), Bounded sequences, Standard limits ( n-th roots, n!/n^n, ln n / n ), and occasionally Algebra of limits (sum, product, quotient).

Worked examples

Cell A — Bounded oscillator over a grower

Cell B — Sign-changing middle (one wall is NOT enough)

Worked example Ex 2 —

$b_{n} = \frac{( - 1 ) ^{n} n}{n + 1}$ Forecast: the sign flips every step. Will it settle, and to what?

Absolute value first. Why this step? When the middle changes sign, an upper bound alone can't trap it (the parent's mistake #2). Bound the size (using $n + 1 > 0$ since $n \geq 1$ ): $∣ b_{n} ∣ = \frac{n}{n + 1} .$
Sandwich the size. Why? $∣ b_{n} ∣ \geq 0$ always, and $\frac{n}{n + 1} \leq \frac{n}{n} = \frac{1}{n}$ : $0 \leq ∣ b_{n} ∣ \leq \frac{1}{n} .$
Squeeze the absolute value. Why? $\frac{1}{n} \to 0$ , so $∣ b_{n} ∣ \to 0$ . And $∣ b_{n} ∣ \to 0 ⟺ b_{n} \to 0$ — a sequence whose distance to $0$ vanishes converges to $0$ regardless of sign.

Verify: $n = 100$ : $\frac{10}{101} \approx 0.099$ , sign alternating; $n = 1 0^{4}$ : $\frac{100}{10001} \approx 0.01$ . Shrinking to $0$ .

Cell C — Factorial / product vs power

Cell D — Root of a polynomial (sandwich by known $n$ th roots)

Worked example Ex 4 —

$b_{n} = n 5 n^{2} + 3$ Forecast: an $n$ th root of something growing polynomially. Does the root crush it to $1$ ?

Trap the inside crudely. Why this step? For $n \geq 1$ , $5 n^{2} + 3$ is between $n$ and $8 n^{2}$ (since $n \leq 5 n^{2} + 3$ and $5 n^{2} + 3 \leq 8 n^{2}$ for $n \geq 1$ ). We want walls that are easy $n$ th roots — quantities whose limit we already know from the standard limits: $n n \leq n 5 n^{2} + 3 \leq n 8 n^{2} = 8^{1/ n} \cdot (n n)^{2} .$
Recall the standard limits. Why? $n n \to 1$ and $8^{1/ n} \to 1$ (any fixed $a > 0$ gives $a^{1/ n} \to 1$ ).
Squeeze. Why? Lower wall $n n \to 1$ ; upper wall $8^{1/ n} (n n)^{2} \to 1 \cdot 1^{2} = 1$ . Both hit $1$ .

Look at the figure below: the green lower wall $n n$ and the orange upper wall $n 8 n^{2}$ start far apart (the root of a large number is big for small $n$ ) but both funnel down to the red line at $1$ . The blue middle $n 5 n^{2} + 3$ has nowhere to escape — it is dragged to $1$ with them. That funnel is the squeeze.

Verify: $n = 10$ : $10503 \approx 1.86$ ; $n = 1000$ : $1000 5 \cdot 1 0^{6} + 3 \approx 1.0155$ . Creeping to $1$ .

Cell E — Sum whose term count grows

Cell F — Degenerate walls (equal-limit check saves you)

Cell G — Real-world word problem

Worked example Ex 7 — sensor noise averaging out

A thermometer reports temperature $T_{n} = 20 + \frac{r _{n}}{n}$ °C on reading $n$ (with $n \in N$ , $n \geq 1$ ), where the noise $r_{n}$ is any value in $[- 3, 3]$ (never larger in magnitude). Does the reported temperature settle, and to what?

Forecast: the noise is bounded, divided by a growing count of readings. Guess the settling value.

Model the noise as bounded. Why this step? We're told $∣ r_{n} ∣ \leq 3$ , i.e. $- 3 \leq r_{n} \leq 3$ — a bounded middle we can wall.
Divide by $n > 0$ and add $20$ . Why? Since $n \geq 1$ , dividing by positive $n$ preserves direction: $20 - \frac{3}{n} \leq T_{n} \leq 20 + \frac{3}{n} .$
Squeeze. Why? Both walls $\to 20$ (constant $\pm$ constant-over-grower). Same limit $\Rightarrow T_{n} \to 20$ .

Verify (units & sanity): every term is in °C; at $n = 60$ readings the band is $20 \pm 0.05$ °C. Physically the noise averages away — the true reading is $20 °C$ .

Cell H — Exam twist (walls that only look different)

Worked example Ex 8 —

$b_{n} = \frac{n sin n + n ^{2}}{n ^{2} + cos n}$ Forecast: two oscillators, top and bottom both blow up. Looks scary — walls seem to fight. Guess anyway.

Bound the oscillators in top and bottom. Why this step? $- n \leq n sin n \leq n$ and $- 1 \leq cos n \leq 1$ . Replace them by their worst cases to build walls. Smallest the fraction can be uses smallest top, largest bottom; largest uses largest top, smallest bottom: $\frac{n ^{2} - n}{n ^{2} + 1} \leq b_{n} \leq \frac{n ^{2} + n}{n ^{2} - 1} (n \geq 2, bottom > 0) .$ Why $n \geq 2$ ? We need $n^{2} + cos n > 0$ ; since $cos n \geq - 1$ , $n^{2} - 1 > 0$ for $n \geq 2$ (recall $n \in N$ ). This is the "eventually" clause — finitely many early terms don't matter.
Simplify each wall. Why? Divide top and bottom by $n^{2}$ : $\frac{1 - \frac{1}{n}}{1 + \frac{1}{n ^{2}}} \to 1, \frac{1 + \frac{1}{n}}{1 - \frac{1}{n ^{2}}} \to 1.$ The walls that looked different both collapse to $1$ .
Squeeze. Both walls $\to 1 \Rightarrow b_{n} \to 1$ .

Verify: $n = 1000$ : numerator $\approx 1 0^{6} \pm 1000$ , denominator $\approx 1 0^{6} \pm 1$ , ratio $\approx 1.00 \pm 0.001$ . Limit $= 1$ .

Active recall

Recall Match the cell to the strategy

Oscillator over a grower ( $\frac{3 c o s ( 5 n ) + 2}{n + 4}$ , Ex 1)? → Cell A: $\pm$ bound over the grower, limit $0$ .
Sign flips every step ( $\frac{( - 1 ) ^{n} n}{n + 1}$ )? → Cell B: squeeze $∣ b_{n} ∣$ , limit $0$ .
$\frac{2 ^{n}}{n !}$ ? → Cell C: bound above by $\frac{4}{n}$ , below by $0$ , limit $0$ .
$n 5 n^{2} + 3$ ? → Cell D: sandwich by known $n$ th roots, limit $1$ .
Walls go to $0$ and $1$ ? → Cell F: theorem says nothing; check for actual divergence.

Recall One-line takeaways (prompt ::: answer)

Why did Ex 6 diverge? ::: The two walls had different limits ( $0$ and $1$ ), so squeeze fails — and the terms genuinely alternate $0, 1$ . In Ex 2 why bound $∣ b_{n} ∣$ not $b_{n}$ ? ::: The middle changes sign; only a two-sided (absolute-value) wall traps both directions. In Ex 8 why require $n \geq 2$ ? ::: To keep the denominator positive; the squeeze only needs to hold eventually. Real-world Ex 7 limit? ::: $20$ °C — bounded noise divided by growing $n$ vanishes. What is the domain of the index $n$ on this page? ::: $n \in N$ , i.e. $n = 1, 2, 3, \dots$ — used to guarantee positive denominators.

Connections

Squeeze theorem for sequences — the parent: statement + $ε$ – $N$ proof.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — the walls in Cells C & D come from here.
Bounded sequences — every oscillator wall ( $∣ sin ∣, ∣ cos ∣, ∣ (- 1)^{n} ∣ \leq 1$ ).
Algebra of limits (sum, product, quotient) — used to simplify the walls (Cells E, H).
Squeeze theorem for functions — same idea on the real line.