Squeeze theorem for sequences: Exercises — Maths deep dive

Definition Ground rules used on this whole page

The index $n$ runs through the positive whole numbers $n = 1, 2, 3, \dots$ (the natural numbers $N$ ). So $n$ is always a positive integer — in particular $n > 0$ , which is why we may divide by $n$ without flipping any inequality. Likewise $n^{2}, n^{2} + 1, n^{2} + n$ are all positive.
A wall is one of the two outer sequences that traps our target; "lower wall" $a_{n}$ sits below, "upper wall" $c_{n}$ sits above.
$N_{0}$ is a threshold index: a specific whole number past which our sandwich $a_{n} \leq b_{n} \leq c_{n}$ is guaranteed to hold. Before $N_{0}$ we make no promises. This is the "eventually" idea: a limit only cares about the tail, so a finite messy start is harmless. In the full $ε$ – $N$ proof (see Limit of a sequence (epsilon-N definition)) one takes $N = max (N_{0}, N_{1}, N_{2})$ so all three facts hold at once.

Level 1 — Recognition

Problem 1.1

For $b_{n} = \frac{cos n}{n}$ , which pair of walls $(a_{n}, c_{n})$ lets you squeeze, and what is the limit?

Recall Solution 1.1

WHAT bounds the messy part. The cosine of any real number lives in $[- 1, 1]$ : that is, $- 1 \leq cos n \leq 1$ . This is the only fact we know about $cos n$ — it never settles down, but it never escapes $[- 1, 1]$ either.

WHY divide by $n$ . We want $\frac{c o s n}{n}$ , so divide all three parts by $n$ . Since $n$ is a positive integer ( $n \geq 1$ ), dividing by it never flips $\leq$ : $- \frac{1}{n} \leq \frac{c o s n}{n} \leq \frac{1}{n} .$ Walls: $a_{n} = - \frac{1}{n}$ , $c_{n} = + \frac{1}{n}$ . Both $\to 0$ .

Conclusion: same limit on both walls $\Rightarrow n \to \infty lim \frac{cos n}{n} = 0.$

Problem 1.2

True or false: if $0 \leq b_{n} \leq \frac{1}{n}$ for all $n \geq 5$ (but the bound fails for $n = 1, 2, 3, 4$ ), we may still conclude $b_{n} \to 0$ .

Recall Solution 1.2

TRUE. A limit is a statement about the tail of a sequence. Changing or ignoring finitely many early terms cannot change where the sequence is heading. Here the threshold index is $N_{0} = 5$ : the squeeze holds "eventually," which is exactly what the theorem requires. See the parent's remark: the bounds only need to hold eventually.

Problem 1.3

Which of these can be squeezed to a limit as stated, and which cannot? (a) $- \frac{1}{n} \leq b_{n} \leq \frac{1}{n}$ (b) $\frac{1}{n} \leq b_{n} \leq 5$ (c) $2 - \frac{1}{n} \leq b_{n} \leq 2 + \frac{1}{n}$ .

Recall Solution 1.3

(a) Both walls $\to 0$ . Squeeze works, $b_{n} \to 0$ .
(b) Lower wall $\frac{1}{n} \to 0$ , upper wall $\to 5$ . Different limits — squeeze says nothing. $b_{n}$ could hover anywhere in $(0, 5]$ .
(c) Both walls $\to 2$ . Squeeze works, $b_{n} \to 2$ .

Level 2 — Application

Problem 2.1

Find $n \to \infty lim \frac{( - 1 ) ^{n}}{n ^{2} + 1}$ .

Recall Solution 2.1

WHAT wiggles. $(- 1)^{n}$ flips between $- 1$ and $+ 1$ forever, so $- 1 \leq (- 1)^{n} \leq 1$ .

WHY divide by $n^{2} + 1$ . Since $n$ is a positive integer, $n^{2} + 1$ is positive for every $n$ , so dividing keeps directions: $- \frac{1}{n ^{2} + 1} \leq \frac{( - 1 ) ^{n}}{n ^{2} + 1} \leq \frac{1}{n ^{2} + 1} .$ Both walls $\to 0$ (denominator blows up). Answer: $0$ .

Problem 2.2

Find $n \to \infty lim \frac{3 + 2 sin ( n ^{3} )}{n}$ .

Recall Solution 2.2

WHAT to bound. $sin (n^{3})$ is unpredictable but $- 1 \leq sin (n^{3}) \leq 1$ , so $2 sin (n^{3}) \in [- 2, 2]$ , hence the numerator $3 + 2 sin (n^{3}) \in [1, 5]$ .

WHY these constant walls help. Divide by $n > 0$ : $\frac{1}{n} \leq \frac{3 + 2 s i n ( n ^{3} )}{n} \leq \frac{5}{n} .$ Lower wall $\frac{1}{n} \to 0$ , upper wall $\frac{5}{n} \to 0$ . WHY does $\frac{5}{n} \to 0$ ? A fixed constant times a sequence going to $0$ still goes to $0$ : given any target band of half-width $ε$ , once $\frac{1}{n} < \frac{ε}{5}$ we have $\frac{5}{n} < ε$ . (This is the product rule of Algebra of limits (sum, product, quotient).) Answer: $0$ .

Problem 2.3

Find $n \to \infty lim \frac{n + cos n}{n ^{2}}$ .

Recall Solution 2.3

Split the numerator. $\frac{n + c o s n}{n ^{2}} = \frac{n}{n ^{2}} + \frac{c o s n}{n ^{2}} = \frac{1}{n} + \frac{c o s n}{n ^{2}}$ .

Bound only the wiggly piece. $- \frac{1}{n ^{2}} \leq \frac{c o s n}{n ^{2}} \leq \frac{1}{n ^{2}}$ , so $\frac{1}{n} - \frac{1}{n ^{2}} \leq \frac{n + c o s n}{n ^{2}} \leq \frac{1}{n} + \frac{1}{n ^{2}} .$ Both walls $\to 0 + 0 = 0$ . Answer: $0$ .

Level 3 — Analysis

Problem 3.1

Find $n \to \infty lim k = 1 \sum n \frac{1}{n ^{2} + k}$ .

Recall Solution 3.1

WHY bound each denominator. There are $n$ terms, index $k$ running $1$ to $n$ . For each such $k$ : $n^{2} + 1 \leq n^{2} + k \leq n^{2} + n$ . Bigger denominator $\Rightarrow$ smaller fraction, so $\frac{1}{n ^{2} + n} \leq \frac{1}{n ^{2} + k} \leq \frac{1}{n ^{2} + 1} .$ Sum over $k$ (all $n$ terms bounded identically): $\frac{n}{n ^{2} + n} \leq \sum_{k = 1}^{n} \frac{1}{n ^{2} + k} \leq \frac{n}{n ^{2} + 1} .$ Limits of walls: $\frac{n}{n ^{2} + n} = \frac{1}{n + 1} \to 0$ , and $\frac{n}{n ^{2} + 1} \to 0$ . Answer: $0$ .

(Contrast with the parent's Example 4, which had $\frac{n}{n ^{2} + k}$ and limit $1$ — the extra factor of $n$ in the numerator there changes everything.)

Problem 3.2

Find $n \to \infty lim n 2^{n} + 3^{n}$ .

Recall Solution 3.2

Squeeze the dominant term. Since $3^{n} \leq 2^{n} + 3^{n} \leq 3^{n} + 3^{n} = 2 \cdot 3^{n}$ , take $n$ th roots. WHY may we? The $n$ th-root map $t \mapsto n t$ is increasing on $t \geq 0$ : if $t_{1} \leq t_{2}$ then $n t_{1} \leq n t_{2}$ (raising the smaller number to the same positive power keeps it smaller). So the inequality directions survive: $n 3^{n} \leq n 2^{n} + 3^{n} \leq n 2 \cdot 3^{n} .$ That is $3 \leq n 2^{n} + 3^{n} \leq 3 \cdot 2^{1/ n}$ .

WHY $2^{1/ n} \to 1$ . This is a standard limit: for any fixed $a > 0$ , $n a \to 1$ . Intuition: to have $2 = x^{n}$ with $n$ huge, $x$ must sit just barely above $1$ — a number even slightly bigger than $1$ , raised to a large power, overshoots $2$ . So the upper wall $\to 3 \cdot 1 = 3$ , the lower wall is constant $3$ . Answer: $3$ (the largest base wins).

Problem 3.3

Show that $n \to \infty lim \frac{⌊ n x ⌋}{n} = x$ for any fixed real $x$ , where $⌊ \cdot ⌋$ is the floor (largest integer $\leq \cdot$ ).

Recall Solution 3.3

WHAT floor guarantees. By definition of floor, for any real $y$ : $y - 1 < ⌊ y ⌋ \leq y$ . Put $y = n x$ : $n x - 1 < ⌊ n x ⌋ \leq n x .$ WHY divide by $n > 0$ . $n$ is a positive integer, so directions are preserved: $x - \frac{1}{n} < \frac{⌊ n x ⌋}{n} \leq x .$ Walls: lower $x - \frac{1}{n} \to x$ , upper the constant $x$ . Both $\to x$ . Answer: $x$ . $■$

Level 4 — Synthesis

Problem 4.1

Find $n \to \infty lim n 3 n^{2} + 5$ .

Recall Solution 4.1

Trap the polynomial between powers of a constant. We build a sandwich valid from an explicit threshold index $N_{0} = 8$ .

Lower side (holds for all $n \geq 1$ ): clearly $3 n^{2} + 5 \geq 1$ .
Upper side: we claim $3 n^{2} + 5 \leq n^{3}$ for all $n \geq 8$ . Check: $n^{3} - 3 n^{2} - 5 = n^{2} (n - 3) - 5$ ; for $n \geq 8$ we have $n^{2} (n - 3) \geq 64 \cdot 5 = 320 > 5$ , so $n^{3} - 3 n^{2} - 5 > 0$ . Thus the bound holds precisely from $N_{0} = 8$ onward.

So for all $n \geq N_{0} = 8$ : $1 \leq 3 n^{2} + 5 \leq n^{3} .$ Take $n$ th roots (increasing map, directions hold): $1 \leq n 3 n^{2} + 5 \leq n n^{3} = (n n)^{3} .$

WHY $n n \to 1$ . Standard limit (proved in the parent's Example 3 by taking logs: $\frac{l n n}{n} \to 0$ ). So the upper wall $\to 1^{3} = 1$ , lower wall constant $1$ . Since the sandwich holds for all $n \geq N_{0}$ , squeeze applies. Answer: $1$ .

Problem 4.2

Find $n \to \infty lim \frac{n !}{n ^{n}} cos (n!)$ . Refer to the figure.

Recall Solution 4.2

WHAT is dangerous. $cos (n!)$ oscillates and $\frac{n !}{n ^{n}}$ is a positive shrinking thing. Product of "wiggle" and "shrink."

Bound the wiggle, ride the shrink. Since $- 1 \leq cos (n!) \leq 1$ and $\frac{n !}{n ^{n}} > 0$ (positive!), multiplying preserves directions: $- \frac{n !}{n ^{n}} \leq \frac{n !}{n ^{n}} cos (n!) \leq \frac{n !}{n ^{n}} .$ WHY both walls die. From the parent (Example 2), $\frac{n !}{n ^{n}} \leq \frac{1}{n} \to 0$ , so $- \frac{n !}{n ^{n}} \to 0$ and $\frac{n !}{n ^{n}} \to 0$ . In the figure this is the yellow band collapsing onto the blue line; the pink sequence, trapped inside, is forced to $0$ as well. Answer: $0$ .

Problem 4.3

Find $n \to \infty lim \frac{1}{n} k = 1 \sum n sin (\frac{k}{n ^{2}})$ .

Recall Solution 4.3

Use the tame bound on sine. For $x \geq 0$ : $0 \leq sin x \leq x$ (sine never overtakes its input on $[0, \infty)$ ; and $sin x \geq 0$ here because each $\frac{k}{n ^{2}} \leq \frac{n}{n ^{2}} = \frac{1}{n}$ is a small non-negative angle). So for each term $0 \leq sin (\frac{k}{n ^{2}}) \leq \frac{k}{n ^{2}} .$ Sum then divide by $n$ : $0 \leq \frac{1}{n} \sum_{k = 1}^{n} sin (\frac{k}{n ^{2}}) \leq \frac{1}{n} \sum_{k = 1}^{n} \frac{k}{n ^{2}} = \frac{1}{n} \cdot \frac{1}{n ^{2}} \cdot \frac{n ( n + 1 )}{2} = \frac{n + 1}{2 n ^{2}} .$ Upper wall $\frac{n + 1}{2 n ^{2}} \to 0$ , lower wall $0$ . Answer: $0$ .

Level 5 — Mastery

Problem 5.1

A student writes: " $0 \leq b_{n} \leq c_{n}$ and $c_{n} \to 3$ , so $b_{n} \to 3$ by squeeze." Explain precisely why this is wrong, and give an explicit $b_{n}$ obeying the hypotheses whose limit is not $3$ .

Recall Solution 5.1

The flaw: the lower wall is the constant $0$ , whose limit is $0 \neq = 3$ . Squeeze requires both walls to share the same limit $L$ . Here the walls disagree ( $0$ vs $3$ ), so the theorem is silent.

Counterexample. Let $c_{n} = 3$ (so $c_{n} \to 3$ ) and $b_{n} = 0$ for all $n$ . Then $0 \leq b_{n} \leq c_{n}$ holds, yet $b_{n} \to 0 \neq = 3$ . The wrong "conclusion" is refuted. $■$

Problem 5.2

Prove: if $∣ b_{n} - L ∣ \leq c_{n}$ for all $n \geq N_{0}$ and $c_{n} \to 0$ , then $b_{n} \to L$ . (This "absolute-value squeeze" is the workhorse form.)

Recall Solution 5.2

Turn the absolute value into two walls. The statement $∣ b_{n} - L ∣ \leq c_{n}$ means exactly $- c_{n} \leq b_{n} - L \leq c_{n} ⟹ L - c_{n} \leq b_{n} \leq L + c_{n} .$ Apply squeeze. Lower wall $L - c_{n} \to L - 0 = L$ ; upper wall $L + c_{n} \to L + 0 = L$ . Same limit $L$ , so $b_{n} \to L$ . $■$

(This is why "show $∣ b_{n} - L ∣ \to 0$ " is the standard route: it is squeeze in disguise, resting on the $\varepsilon$–$N$ definition.)

Problem 5.3

Find $n \to \infty lim \frac{2 ^{n}}{n !}$ using a squeeze. (No ratio test.)

Recall Solution 5.3

Write out the product and isolate a fixed head. For $n \geq 2$ : $\frac{2 ^{n}}{n !} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \dots \frac{2}{n} .$ Keep the last factor $\frac{2}{n}$ and bound the middle factors by $1$ : $0 < \frac{2 ^{n}}{n !} = 2 \cdot (\frac{2}{3} \cdot \frac{2}{4} \dots \frac{2}{n - 1}) \cdot \frac{2}{n} \leq 2 \cdot 1 \cdot \frac{2}{n} = \frac{4}{n} .$ (Each factor $\frac{2}{3}, \dots, \frac{2}{n - 1}$ is $\leq 1$ .) Squeeze: $0 < \frac{2 ^{n}}{n !} \leq \frac{4}{n} \to 0$ . Answer: $0$ .

Problem 5.4

Give a sequence $(b_{n})$ that is bounded ( $- 1 \leq b_{n} \leq 1$ ) but cannot be assigned a limit by squeeze — and say why, connecting to convergence.

Recall Solution 5.4

Take $b_{n} = (- 1)^{n}$ . It sits inside the walls $- 1 \leq b_{n} \leq 1$ , but those walls converge to different numbers ( $- 1$ and $+ 1$ ), so squeeze gives nothing. Deeper reason: $(- 1)^{n}$ genuinely diverges — its terms cluster at both $- 1$ and $+ 1$ forever, so no squeeze could ever pin it (there is nothing to pin it to). Boundedness alone does not force convergence; you need either matching walls (squeeze) or monotonicity (see Monotone convergence theorem).

Recall the whole page

Recall Rapid review (cover the right side)

Walls for $\frac{c o s n}{n}$ (P1.1) ::: $- \frac{1}{n}$ and $+ \frac{1}{n}$ , both $\to 0$ , so limit $0$ . Does $\frac{1}{n} \leq b_{n} \leq 5$ pin a limit? (P1.3b) ::: No — the walls have different limits, so squeeze is silent. $lim \frac{( - 1 ) ^{n}}{n ^{2} + 1}$ (P2.1) ::: $0$ , walls $\pm \frac{1}{n ^{2} + 1}$ . $lim \frac{3 + 2 s i n ( n ^{3} )}{n}$ (P2.2) ::: $0$ , walls $\frac{1}{n}$ and $\frac{5}{n}$ . $lim \frac{n + c o s n}{n ^{2}}$ (P2.3) ::: $0$ , walls $\frac{1}{n} \pm \frac{1}{n ^{2}}$ . $lim \sum_{k = 1}^{n} \frac{1}{n ^{2} + k}$ (P3.1) ::: $0$ , walls $\frac{n}{n ^{2} + n}$ and $\frac{n}{n ^{2} + 1}$ . $lim n 2^{n} + 3^{n}$ (P3.2) ::: $3$ (the largest base dominates). $lim \frac{⌊ n x ⌋}{n}$ (P3.3) ::: $x$ , via $x - \frac{1}{n} < \frac{⌊ n x ⌋}{n} \leq x$ . $lim n 3 n^{2} + 5$ (P4.1) ::: $1$ , trapped between $1$ and $(n n)^{3}$ for $n \geq N_{0} = 8$ . $lim \frac{n !}{n ^{n}} cos (n!)$ (P4.2) ::: $0$ , envelope $\pm \frac{n !}{n ^{n}}$ . $lim \frac{1}{n} \sum_{k = 1}^{n} sin (k / n^{2})$ (P4.3) ::: $0$ , upper wall $\frac{n + 1}{2 n ^{2}}$ . Absolute-value squeeze (P5.2) ::: $∣ b_{n} - L ∣ \leq c_{n}$ with $c_{n} \to 0$ gives $b_{n} \to L$ . $lim \frac{2 ^{n}}{n !}$ (P5.3) ::: $0$ (bounded above by $\frac{4}{n}$ ). Why $(- 1)^{n}$ resists squeeze (P5.4) ::: its two walls $\pm 1$ have different limits; it diverges. One-line moral of the whole page ::: two walls with the SAME limit force the trapped sequence to that limit; different-limit walls prove nothing.

Connections

Parent: Squeeze theorem — statement, proof, base examples.
Limit of a sequence (epsilon-N definition) — the machinery behind every " $\to 0$ ."
Bounded sequences — where the walls $\pm 1$ come from.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — $n a \to 1$ , $\frac{n !}{n ^{n}} \to 0$ used above.
Monotone convergence theorem — the other route to a limit when walls don't match.
Squeeze theorem for functions — the continuous cousin (used for $0 \leq sin x \leq x$ ).