📝 Squeeze theorem for sequences (35 MCQs)

📖 Explanation: The Squeezing Theorem requires that both the lower bounding sequence and the upper bounding sequence converge to the exact same limit $L$. In this student's reasoning, the bounds are constants -1 and 1. Since $\lim_{n \to \infty} (-1) = -1$ and $\lim_{n \to \infty} (1) = 1$, and $-1 \neq 1$, the theorem cannot be applied to deduce the limit of the middle sequence. The correct bounds would be $-1/n$ and $1/n$, which both converge to 0.

📖 Explanation: To squeeze $n!/n^n$, we expand it as $\frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n}{n}$. Every factor except the first is less than or equal to 1. Thus, the entire product is bounded above by $\frac{1}{n} \cdot 1 \cdot 1 \cdots = \frac{1}{n}$. Since $b_n > 0$ for all $n$, we have $0 \leq b_n \leq \frac{1}{n}$. As $n \to \infty$, both 0 and $1/n$ approach 0, confirming the limit is 0 via the Squeezing Theorem. This avoids complex calculus tools.

📖 Explanation: The definition of the limit of a sequence concerns the behavior as $n \to \infty$. Finite initial terms, or even a finite number of undefined terms, do not affect the existence or value of the limit. The Squeezing Theorem explicitly allows for inequalities to hold 'eventually' (i.e., for all $n$ beyond some index $N$). Since the bounding sequences both converge to 5 for large $n$, the middle sequence must also converge to 5 regardless of its early behavior.

📖 Explanation: The Squeezing Theorem applies to sequences based on their values at integer indices. Even if the bounding functions $f(x)$ and $g(x)$ are continuous, we only evaluate them at integer points $n$ to create bounding sequences $f(n)$ and $g(n)$. If these evaluated sequences satisfy the inequality and share a common limit $L$, the theorem guarantees $a_n \to L$. The continuous nature of the envelope functions helps visualize the 'squeezing' effect but the rigorous justification relies on the discrete evaluation at integers.

📖 Explanation: Standard algebraic limit laws require the individual limits to exist. For $\frac{(-1)^n}{\sqrt{n}}$, the term $(-1)^n$ has no limit, making the quotient law inapplicable directly. While one could use absolute convergence, the Squeezing Theorem provides a direct path: $-\frac{1}{\sqrt{n}} \leq a_n \leq \frac{1}{\sqrt{n}}$. Both bounds go to 0, forcing the oscillating sequence to 0. This highlights the theorem's utility for handling bounded oscillation combined with decay, where pure algebra fails.

📖 Explanation: Convergence to a limit $L$ means the terms get arbitrarily close to $L$, not that they become identical to $L$. A classic counterexample is $b_n = L + \frac{(-1)^n}{n}$. This sequence is squeezed between $L - \frac{1}{n}$ and $L + \frac{1}{n}$, both converging to $L$. However, $b_n$ never equals $L$ and continues to oscillate forever. The theorem constrains the range of values, ensuring proximity to the limit, but does not enforce constancy or monotonicity.

📖 Explanation: We can set up the inequality $0 \leq |E_n| \leq \frac{K}{n^p}$. The lower bound is constantly 0. The upper bound $K n^{-p}$ approaches 0 as $n \to \infty$ if and only if the exponent $p$ is strictly positive. If $p = 0$, the bound is constant $K$, providing no squeezing to 0. If $p < 0$, the bound diverges. Therefore, for any $p > 0$, the Squeezing Theorem guarantees $|E_n| \to 0$, implying the algorithm's error vanishes asymptotically.

📖 Explanation: Steps 1 through 4 constitute a perfectly valid application of the Squeezing Theorem to prove that the limit is 1. The error lies exclusively in Step 5. Convergence to a limit does not imply that the sequence becomes equal to that limit. The sequence $\frac{n + \cos(n)}{n} = 1 + \frac{\cos(n)}{n}$ fluctuates around 1 for every finite $n$; it is never exactly 1 (except potentially at isolated points). This distractor tests the distinction between asymptotic behavior and exact equality.

📖 Explanation: Directly bounding by constants like 3 and 5 is insufficient as they don't share a limit. Recognizing $5^n$ as dominant, we write $5^n < 3^n + 5^n < 5^n + 5^n = 2 \cdot 5^n$. Taking the nth root gives $5 < a_n < 5 \cdot 2^{1/n}$. As $n \to \infty$, $2^{1/n} \to 1$, so the upper bound approaches 5. The lower bound is constantly 5. Both converge to 5, squeezing $a_n$ to 5. This demonstrates identifying asymptotic dominance to construct tight, convergent bounds rather than loose static ones.

📖 Explanation: Limits describe long-term behavior. Many meaningful sequences may be undefined at $n=0$ (e.g., involving $1/n$) or exhibit irregular behavior initially before settling into a pattern. Requiring the inequality to hold for all $n \geq 1$ would unnecessarily exclude such sequences. By allowing the condition to hold eventually (for $n \geq N$), the theorem remains robust and applicable to virtually all convergent scenarios, reinforcing that finite prefixes are irrelevant to infinite limits.

📖 Explanation: The Squeezing Theorem relies solely on the magnitude of the oscillating component being bounded. Whether $\sin(n)$, $\sin(n^2)$, or $\sin(e^n)$ appears, the range is always $[-1, 1]$. Thus $-1/n \leq d_n \leq 1/n$ holds regardless of oscillation frequency. The rate of oscillation affects how quickly individual terms fluctuate but has zero bearing on the asymptotic envelope created by the $1/n$ factor. This highlights that squeezing addresses amplitude decay, not frequency characteristics.

📖 Explanation: While $0 \leq a_n \leq \frac{n+1}{n} \to 1$ is technically true, it fails to squeeze the sequence to a unique value because the bounds converge to different limits (0 and 1). The actual limit is $\ln 2 \approx 0.693$. Effective squeezing requires both bounds to meet at the true limit. This problem illustrates that finding valid inequalities is not enough; they must be sharp enough to coincide asymptotically. Better bounds would involve integrals or recognizing the Riemann sum structure.

📖 Explanation: This is a specific corollary often derived directly from the Squeezing Theorem. Since $-|x| \leq x \leq |x|$ for any real number, substituting $a_n$ gives $-|a_n| \leq a_n \leq |a_n|$. If $|a_n| \to 0$, then $-|a_n| \to 0$ as well. Both outer sequences converge to 0, so by the Squeezing Theorem, $a_n$ must also converge to 0. This connects the concept of absolute convergence to signed convergence rigorously.

📖 Explanation: For large $n$, $n^2 + 3n < 2n^2$ and $n^3 - n > n^3/2$ are both true. Thus the inequality holds eventually. The upper bound simplifies to $4/n$, which approaches 0. The lower bound is 0 (since terms are positive for $n>1$). Since both bounds converge to 0, the squeeze is valid. The key insight is verifying that the simplified bounds are indeed valid overestimates/underestimates asymptotically, not just algebraically similar.

📖 Explanation: These theorems address fundamentally different concepts. The Squeezing Theorem (for sequences) establishes convergence to a limit $L$ by trapping terms between two other sequences converging to $L$. It says nothing about attaining specific values. The Intermediate Value Theorem guarantees that a continuous function on $[a,b]$ takes every value between $f(a)$ and $f(b)$. Confusing them is a common conceptual error; one is about asymptotic confinement, the other about surjectivity on intervals.

📖 Explanation: Even though the floor function is discontinuous, it satisfies strict algebraic inequalities: $n-1 < \lfloor n \rfloor \leq n$ for all real $n$. Dividing by $n$ (for $n>0$) gives $1 - \frac{1}{n} < a_n \leq 1$. As $n \to \infty$, $1 - 1/n \to 1$ and the upper bound is constantly 1. Thus, $a_n \to 1$. This demonstrates that the Squeezing Theorem works perfectly with discrete/discontinuous components as long as analytic bounds exist.

📖 Explanation: The sine function takes negative values. Therefore, $\frac{\sin n}{n}$ is frequently negative, violating the stated lower bound of 0. While the conclusion (limit is 0) happens to be correct, the justification is flawed. The correct inequality is $-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}$ or using absolute values $0 \leq |\frac{\sin n}{n}| \leq \frac{1}{n}$. Precision in establishing valid bounds for all $n$ is critical for rigorous application of the theorem.

📖 Explanation: Rearranging $|x_n - L| \leq \epsilon_n$ yields $-\epsilon_n \leq x_n - L \leq \epsilon_n$, or $L - \epsilon_n \leq x_n \leq L + \epsilon_n$. Define $a_n = L - \epsilon_n$ and $c_n = L + \epsilon_n$. Since $\epsilon_n \to 0$, both $a_n \to L$ and $c_n \to L$. This is precisely the hypothesis of the Squeezing Theorem. This connection shows that error bounds in numerical methods are practically applications of squeezing, providing theoretical guarantees for computational accuracy.

📖 Explanation: The Squeezing Theorem is a sufficient condition for convergence, not a tool for proving divergence directly. To apply it, you need bounding sequences sharing a limit. For $(-1)^n$, any valid lower bound must be $\leq -1$ infinitely often and any upper bound $\geq 1$ infinitely often. No pair of sequences converging to the same $L$ can satisfy this. The inability to construct such bounds reflects the sequence's intrinsic divergence, but the theorem itself doesn't output 'diverges'; it simply fails to apply.

📖 Explanation: There are $n$ terms. Each term is between $\frac{1}{\sqrt{n^2+n}}$ and $\frac{1}{\sqrt{n^2+1}}$. Summing gives $\frac{n}{\sqrt{n^2+n}} \leq a_n \leq \frac{n}{\sqrt{n^2+1}}$. Simplifying: lower bound is $\frac{1}{\sqrt{1+1/n}} \to 1$; upper bound is $\frac{1}{\sqrt{1+1/n^2}} \to 1$. Both converge to 1, so $a_n \to 1$. This technique of bounding sums by (number of terms) × (min/max term) is a classic and powerful application of squeezing for sequences defined as sums.

📖 Explanation: The standard Squeezing Theorem relies on the order property of real numbers ($\leq$). Complex numbers do not form an ordered field; you cannot say $z_1 \leq z_2$. For complex sequences, one uses the modulus: if $|b_n - L| \leq \epsilon_n$ with $\epsilon_n \to 0$ (real), then $b_n \to L$. This is effectively squeezing the real-valued distance $|b_n - L|$ to 0. Recognizing domain restrictions prevents misapplication of real-analysis theorems to complex contexts.

📖 Explanation: The left bound $3 - 1/n \to 3$. The right bound $3 + 1/n^2 \to 3$. Despite differing convergence rates (one is $O(1/n)$, the other $O(1/n^2)$), both limits are identically 3. The Squeezing Theorem requires only that the limits match, not that the approach speeds be symmetric or identical. This reinforces that asymptotic destination matters, not trajectory shape.

📖 Explanation: Note that $n^2 + 1 > n^2$, so $\frac{1}{n^2+1} < \frac{1}{n^2}$. Multiplying by positive $n$ preserves this: $\frac{n}{n^2+1} < \frac{n}{n^2}$. Since $\cos(n) \leq 1$, we have $a_n \leq \frac{n}{n^2+1} < \frac{n}{n^2}$. So actually $a_n < \frac{n}{n^2}$ IS true. Wait—re-evaluating: The student wrote $a_n \leq \frac{n}{n^2}$. Since $\frac{n}{n^2+1} < \frac{n}{n^2}$, and $a_n \leq \frac{n}{n^2+1}$, transitivity gives $a_n < \frac{n}{n^2}$. So the inequality is actually correct. However, option B claims it's wrong. Let's reconsider: Perhaps the issue is looseness? No, loose bounds are fine if they converge. Actually, the inequality IS valid. But among choices, if forced to critique, maybe precision? Re-reading carefully: The student's bound is valid and sufficient. But if the question assumes a flaw, perhaps it's that $\frac{n}{n^2} = 1/n$ while actual max is slightly less. Still, 1/n → 0. Given typical exam traps, sometimes 'wrong direction' is the intended answer if students confuse $n^2+1 > n^2 \implies \text{fraction smaller}$. But here smaller upper bound still ≤ larger upper bound. I'll select A as correct reasoning, but since I must pick from provided options and A says 'Yes', that seems right. However, looking at the pattern of error analysis questions, let's assume the prompt expects identification of a subtle issue. Actually, re-evaluating Option B: it says inequality direction is wrong. That's factually incorrect. Option A is fully correct. I will choose A.

📖 Explanation: For $n \sin(n) / n = \sin(n)$, bounds are -1 and 1, which don't converge to same limit. Squeezing fails. For $\sin(n^2)/n$ or $\sin(n)/\sqrt{n}$, bounds still go to 0. Adding constant shifts limit to 1, but squeezing still works with adjusted bounds. Only when the oscillation amplitude doesn't decay (or grows) does squeezing to a point fail. This tests understanding that squeezing requires decaying envelopes.

📖 Explanation: Set $\sqrt[n]{n} = 1 + \delta_n$ with $\delta_n > 0$. Then $n = (1+\delta_n)^n \geq 1 + n\delta_n + \frac{n(n-1)}{2}\delta_n^2$ (binomial theorem, keeping quadratic term). Ignoring linear term: $n > \frac{n(n-1)}{2}\delta_n^2 \implies \delta_n^2 < \frac{2}{n-1} < \frac{2}{n}$ for $n>2$. Thus $\delta_n < \sqrt{2/n}$. This constructs a tight upper bound converging to 1. This derivation showcases advanced algebraic manipulation to create squeezable bounds where simple inspection fails.

📖 Explanation: Visual contact with decaying envelopes $\pm 1/x$ at integer points means $-1/n \leq a_n \leq 1/n$ holds (at least at touchpoints, and presumably between). Since both envelopes → 0, squeezing forces $a_n \to 0$. Alternation isn't guaranteed (could touch same side twice). Monotonicity clearly false. Tangency is visual interpretation, but convergence is the rigorous mathematical consequence guaranteed by the depicted bounding behavior.

📖 Explanation: Without a lower bound, $a_n$ could be $-n$, $-n^2$, etc., while still satisfying $a_n \leq b_n \to 0$. The upper bound controls from above but says nothing about downward behavior. Squeezing requires confinement from BOTH sides to prevent escape in either direction. The non-negativity condition $0 \leq a_n$ serves as the crucial lower bound anchoring the sequence, preventing negative divergence.

📖 Explanation: While $0 \leq \sin(k/n) \leq 1$ implies $0 \leq a_n \leq 1$, these bounds don't coincide. The sequence actually converges to $1 - \cos(1) \approx 0.46$. Squeezing with crude bounds only places it in [0,1]. To find exact limit, recognize Riemann sum structure or use tighter analytical bounds. This illustrates squeezing's limitation: it confirms boundedness/convergence only when bounds are sharp; otherwise, other techniques (integration, asymptotics) are needed for precise values.

📖 Explanation: Rewrite as $3 - \frac{1}{\ln n} \leq a_n \leq 3 + \frac{1}{\ln n}$. Since $\ln n \to \infty$, $1/\ln n \to 0$. Both bounds → 3. Slow growth of log doesn't matter; any unbounded denominator drives fraction to 0. This reinforces that squeezing works with ANY null sequence as envelope, regardless of decay rate, as long as it genuinely approaches 0.

📖 Explanation: Upper bound alone only tells us $\limsup a_n \leq 5$. The sequence could converge to 0, -100, or diverge to $-\infty$. Convergence to specifically 5 requires confinement from below by something also approaching 5. This is the core mechanism of squeezing: bilateral constraint. Unilateral bounds provide only partial information about asymptotic behavior.

📖 Explanation: Mathematically, this is squeezing to $P^*$. Biologically, it means deviations from equilibrium shrink over time; the system exhibits asymptotic stability toward $P^*$. Unlike permanent oscillation or extinction, the population settles at a predictable steady state. This connects abstract convergence to tangible ecological resilience, showing how squeezing formalizes 'returning to equilibrium' in dynamic systems.

📖 Explanation: First three have decaying amplitudes: bounded numerator / growing denominator → 0. Last one oscillates between -1 and 1 forever; no decaying envelope exists. Any valid bounds must include ±1, which don't converge to same limit. This is the canonical example of a bounded divergent sequence where squeezing to 0 is impossible.

📖 Explanation: Although $(-1)^n$ alternates, it's always in [-1,1]. So $n^2 - n \leq \text{numerator} \leq n^2 + n$. Dividing by $n^2+1$ gives bounds both → 1. Parity-specific analysis is redundant when uniform bounds suffice. This efficiency is squeezing's power: handle oscillation globally via magnitude bounds rather than case-by-case enumeration.

📖 Explanation: Sequences are functions on integers. If undefined at finitely many points, limit unaffected. If undefined at infinitely many points BUT defined for all $n \geq N$, still fine. Only if undefined arbitrarily far out (no tail) is limit meaningless. For $\sin(n\pi)/n$, it's actually 0 at all integers (since sin(nπ)=0), so no issue. But hypothetically, eventual definition suffices. This clarifies domain requirements.

📖 Explanation: Symmetry irrelevant. Left → 2, right → 2. Squeeze complete. Numerical suggestion motivated the bounds, but analytical bounds constitute proof. This validates asymmetric convergence rates in squeezing arguments.