📝 Eventually properties of sequences (31 MCQs)

📖 Explanation: Although the specific value $a_5 = 100$ creates a significant disruption in the pattern, the definition of properties holding eventually allows us to ignore finitely many initial terms. For $n > 5$, the expression simplifies to $n-5$, which is clearly strictly increasing. Therefore, despite the anomaly at $n=5$, the sequence possesses the property of being eventually strictly increasing because the tail behaves monotonically.

📖 Explanation: Convergence depends entirely on the tail of the sequence, not its initial behavior. While the sequence is not globally monotone due to the chaotic start, it is eventually decreasing and bounded below by zero. The Monotone Convergence Theorem applies to sequences that are *eventually* monotone. The student's error lies in assuming that global regularity is required for convergence, whereas only the asymptotic behavior determines the existence of a limit.

📖 Explanation: To find when the sequence becomes strictly decreasing, we analyze the ratio $\frac{a_{n+1}}{a_n} = \frac{10}{n+1}$. For the sequence to be strictly decreasing, this ratio must be less than 1. Solving $\frac{10}{n+1} < 1$ yields $n+1 > 10$, or $n > 9$. Thus, the condition holds starting at $n=10$. This demonstrates that even if a sequence increases initially (which this one does for $n < 9$), it can still be classified as eventually strictly decreasing once the factorial growth dominates the exponential growth.

📖 Explanation: Visual interpretation of sequences requires focusing on the long-term trend rather than local anomalies. The graph shows that after $n=51$, the points form a decreasing sequence bounded below by the horizontal axis. According to the theorem on monotone sequences, an eventually decreasing sequence that is bounded below must converge. The sharp drop represents a finite number of irregular terms, which do not affect the existence of the limit, confirming convergence to zero based on the eventual behavior shown in the graph.

📖 Explanation: This problem mixes the concepts of alternating series behavior and eventual properties. Although the first 100 terms alternate, preventing global monotonicity, the definition of 'eventually' permits discarding these terms. For $n > 100$, the sequence is $\frac{1}{n}$, which is strictly decreasing and positive. Since the tail is strictly decreasing and bounded below by 0, the Monotone Convergence Theorem guarantees convergence to 0. The initial oscillation is irrelevant to both the eventual monotonicity and the limit.

📖 Explanation: Monotonicity is determined exclusively by the sign of the difference $a_{n+1} - a_n$, not its magnitude. While the magnitude tells us about the rate of change, the sign dictates direction. Here, the denominator $n^2+1$ is always positive. The numerator $n-100$ is negative for $n < 100$ and positive for $n > 100$. Therefore, despite the difference approaching zero, the sequence is strictly decreasing for the first 100 terms and strictly increasing thereafter. It is eventually strictly increasing because the sign stabilizes to positive.

📖 Explanation: This scenario models a recursive sequence where the defining relation holds only eventually. The initial fluctuations represent a finite number of terms that do not influence the asymptotic limit. Assuming the sequence converges to $L$, we solve $L = \sqrt{L+6}$, yielding $L^2 - L - 6 = 0$, so $L=3$ or $L=-2$. Since population must be non-negative and $P_{21}=3$, the limit is 3. The key insight is that the eventual recursive definition completely overrides the chaotic history for determining the limit.

📖 Explanation: The student's premise that unboundedness precludes monotonicity is false (e.g., $a_n=n$ is unbounded and increasing). However, their conclusion for this specific sequence is correct for the wrong reason. The factor $\sin(n)$ oscillates between -1 and 1 infinitely often. Even though $n^2$ grows large, the sign changes of $\sin(n)$ ensure that $a_{n+1} - a_n$ changes sign infinitely often. Thus, the sequence is NOT eventually monotone, but the reason is the persistent oscillation of the trigonometric term, not the unboundedness itself.

📖 Explanation: This question tests the distinction between global and eventual properties alongside convergence. Both sequences share the same tail ($1/n$), so they must share the same limit (0). Sequence $a_n$ is globally strictly decreasing. Sequence $b_n$ has massive spikes for odd $n < 50$, violating global monotonicity. However, for $n \geq 50$, $b_n$ becomes identical to $a_n$. Therefore, $b_n$ is eventually strictly decreasing. Convergence is identical, but monotonicity differs globally while matching eventually.

📖 Explanation: The Completeness Axiom and the Monotone Convergence Theorem rely on the behavior of the sequence's tail. If the sequence is eventually increasing, there exists some index $N$ after which $a_n \leq M$. The limit $L$ is the least upper bound of the set $\{a_n : n \geq N\}$. Since all terms in this tail are $\leq 10$, the supremum must be $\leq 10$. The fact that earlier terms exceeded 10 is irrelevant to the limit's value, as limits describe asymptotic behavior, not transient excursions.

📖 Explanation: While calculus confirms f&#039;(x) = 1/x - 1 < 0 for $x > 1$, proving eventual monotonicity purely from numerical data is logically insufficient. 'Eventually' implies existence of an $N$, but computation can never verify infinity. However, conceptually, the question highlights the gap between empirical observation and analytical proof. Even though the trend is robust here, in general sequences, subtle transitions can occur beyond computational reach. Rigorous proof requires analyzing the sign of $a_{n+1}-a_n$ or f&#039;(x) analytically, not just observing a million data points.

📖 Explanation: Convergence depends on terms getting arbitrarily close to the limit, regardless of order. Changing infinitely many terms to alternate signs (e.g., $L + (-1)^n/n$) preserves convergence to $L$ but destroys monotonicity because the sign of the difference flips infinitely often. Options A, B, and D preserve both convergence and eventual monotonicity. Option C specifically targets the distinction: convergence allows oscillation within shrinking bounds, whereas monotonicity forbids direction changes. This highlights that convergence is a weaker condition than eventual monotonicity.

📖 Explanation: The definition of a limit states that if $\lim r_n = L$, then for any $\epsilon > 0$, eventually $|r_n - L| < \epsilon$. Choosing $\epsilon$ such that $L + \epsilon < 1$ (possible since $L=0.5$), we guarantee that eventually $r_n < 1$. Since terms are positive, $a_{n+1}/a_n < 1$ implies $a_{n+1} < a_n$. Thus, the sequence is indeed eventually strictly decreasing. This connects the analytic definition of limits directly to the discrete property of monotonicity, validating the student's claim rigorously.

📖 Explanation: This is a classic misconception conflating convergence with monotonicity. While $|\sin(n)/n|$ decreases eventually, the signed sequence $\sin(n)/n$ continues to oscillate between positive and negative values as $n \to \infty$. Because $\sin(n)$ changes sign infinitely often, the sequence values go up and down across zero indefinitely. Therefore, it is never eventually monotone, despite converging to 0. This distinguishes absolute decay from signed monotonic behavior.

📖 Explanation: Calculating differences: $a_{n+1} - a_n = 1 + (-1)^{n+1} - (-1)^n$. If $n$ is even, diff = $1 - 1 - 1 = -1$. If $n$ is odd, diff = $1 + 1 + 1 = 3$. The difference alternates between -1 and 3. Since it takes negative values infinitely often, the sequence is NEVER eventually increasing (neither strictly nor non-strictly). Wait, re-evaluating: $a_1=0, a_2=3, a_3=2, a_4=5...$. It goes up, down, up, down. It is NOT eventually increasing. Correct answer should reflect this. Let me adjust options. Actually, checking standard curriculum examples, $n + (-1)^n$ is often used to show failure of monotonicity. I will select option D as correct in spirit, but looking at provided options, let's re-read carefully. Ah, if the question asks to distinguish, and the sequence fails both, D is correct. However, if I must choose from the provided list where B is a common distractor type, I need to ensure accuracy. Actually, $a_n = n + 2(-1)^n$ would fail too. Let's assume the question intended a sequence like $a_n = n$ for $n>10$ and messy before. But for $n+(-1)^n$, it is indeed neither. I will mark D as correct. Explanation clarifies why 'eventually' doesn't save it here.

📖 Explanation: The definition of 'eventually strictly decreasing' requires the existence of an integer $N$ such that $a_{n+1} < a_n$ for all $n \geq N$. In this scenario, choosing $N=1005$ satisfies the condition perfectly. The behavior prior to $N$, including the brief increase, is irrelevant to the classification. This reinforces that 'eventually' acts as a filter removing finite prefixes, allowing classification based solely on the permanent long-term trend.

📖 Explanation: Analyzing the ratio $\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n+1}{n+1} (\frac{n}{n+1})^n = (1 - \frac{1}{n})^n$. We know this sequence increases to $1/e \approx 0.368$. Since the limit is strictly less than 1, and the sequence of ratios is monotonic, the ratio is always less than 1 for $n \geq 1$. Actually, since it approaches $1/e$ from below, it is ALWAYS < 1. Thus it is strictly decreasing for all $n$. If the limit were, say, 0.9 but approached from above, it would only be *eventually* decreasing. Here, the limit being $<1$ is the sufficient condition for eventual strict decrease, confirming option A.

📖 Explanation: Sequences are functions with domain $\mathbb{Z}^+$. If $a_1$ is undefined, strictly speaking, it's not a sequence on $\mathbb{Z}^+$. However, in analysis contexts involving 'eventually', we often implicitly restrict the domain to where the sequence is well-defined. If the sequence is well-defined for all $n \geq 2$, and satisfies the conditions for $n \geq N \geq 2$, the theorem applies to the subsequence starting at $\max(2, N)$. The limit exists for this valid tail. Option B captures this pragmatic mathematical handling of singularities in eventual analysis.

📖 Explanation: While computing terms provides evidence, it isn't proof. The ratio test is for series convergence, not sequence monotonicity directly (though related). Checking positivity doesn't prove decrease. Analyzing the continuous function $f(x)$ via derivatives is the standard rigorous method. f&#039;(x) = \frac{1}{2\sqrt{x+1}} - \frac{1}{2\sqrt{x}}. Since $\sqrt{x+1} > \sqrt{x}$, the first term is smaller, making f&#039;(x) < 0 for all $x \geq 1$. This proves $f$ is strictly decreasing on $[1, \infty)$, which implies the sequence is strictly decreasing (and thus eventually strictly decreasing). This links continuous calculus tools to discrete sequence properties.

📖 Explanation: This tests the precise definition against common misconceptions. An eventually strictly increasing sequence can be bounded (e.g., $-1/n$ increases to 0). It can have negative terms. Its limit can be finite. The ONLY necessary condition among the choices is the definitional one: existence of a threshold $N$ beyond which strict inequality holds. Options A, B, and D describe properties that some eventually increasing sequences have, but none are necessary. This reinforces the definition over intuitive but incorrect associations.

📖 Explanation: We examine the sign of the difference $d_n = \frac{(-1)^n}{n} + \frac{1}{n^2}$. For even $n$, $d_n > 0$. For odd $n$, $d_n = -\frac{1}{n} + \frac{1}{n^2} = \frac{1-n}{n^2}$, which is negative for all $n > 1$. Since the difference alternates sign infinitely often (positive for even, negative for odd), the sequence increases and decreases perpetually. No matter how large $N$ is chosen, there will always be odd $n > N$ where $d_n < 0$ and even $n > N$ where $d_n > 0$. Thus, it is never eventually monotone, despite the $1/n^2$ term suggesting a drift.

📖 Explanation: The peer misunderstands 'eventually'. Strict decrease requires $a_{n+1} < a_n$. For $n > 100$, $5 + \frac{1}{n+1} < 5 + \frac{1}{n}$ holds true. The behavior for $n \leq 100$ (where $a_{n+1} = a_n$) violates *global* strict monotonicity but is completely irrelevant to *eventual* strict monotonicity. The definition explicitly allows discarding any finite prefix. The refutation rests on isolating the tail where the strict inequality permanently holds.

📖 Explanation: Differentiating $f(x)$: f&#039;(x) = \frac{1 - \ln x}{x^2}. The derivative is negative when $\ln x > 1$, i.e., $x > e \approx 2.718$. Since the sequence is defined on integers, the condition $n > e$ is first satisfied at $n = 3$. At $n=2$, $\ln 2 \approx 0.69 < 1$, so f&#039;(2) > 0 (increasing). Thus, the sequence increases from $n=1$ to $n=2$ and strictly decreases for all $n \geq 3$. This demonstrates translating continuous critical points to discrete eventual thresholds.

📖 Explanation: Mathematical models often assume ideal conditions that hold asymptotically. Real systems have initialization phases, warm-up periods, or transient disturbances that violate theoretical monotonicity. Without the 'eventually' qualifier, the Monotone Convergence Theorem would be inapplicable to almost all empirical data. By permitting the exclusion of finite initial segments, the theorem bridges pure mathematics and applied science, allowing analysts to validate convergence based on stabilized system behavior rather than rejecting models due to unavoidable startup artifacts.

📖 Explanation: This links integral calculus to sequence monotonicity. Geometrically, $a_n$ is the area under $y=1/x$ from $n$ to $n+1$. Since $1/x$ is strictly decreasing for $x>0$, the rectangle/area over $[n+1, n+2]$ must be strictly smaller than the area over $[n, n+1]$. Formally, $a_{n+1} - a_n = \int_{n+1}^{n+2} \frac{dx}{x} - \int_n^{n+1} \frac{dx}{x}$. Shifting indices shows this difference is negative because the integrand decreases. Thus, it is strictly decreasing for all $n \geq 1$, hence eventually strictly decreasing. This uses functional properties to deduce sequential ones.

📖 Explanation: The given condition is $a_{n+1} \leq a_n$, which defines *decreasing* (non-strict). Option D claims *strictly* decreasing ($a_{n+1} < a_n$). A sequence could be constant (e.g., $a_n = 5$ for $n \geq 50$), satisfying the premise but falsifying D. Options A, B, and C are true: bounded below by 0, eventually decreasing by definition, and convergent by Monotone Convergence Theorem. Identifying the false statement requires careful attention to the distinction between strict and non-strict inequalities in definitions.

📖 Explanation: The sum of an increasing and decreasing sequence is indeterminate without knowing relative rates. Example 1: $a_n = 2n, b_n = -n \Rightarrow sum = n$ (increasing). Example 2: $a_n = n, b_n = -2n \Rightarrow sum = -n$ (decreasing). Example 3: $a_n = n + (-1)^n/n, b_n = -n$ (adjusting to maintain monotonicity individually is tricky, but simpler: $a_n=n, b_n=-n+1/n$ for $n \geq 1$; $b$ is eventually decreasing? Derivative of $-x+1/x$ is $-1-1/x^2 < 0$. Sum = $1/n$, decreasing). Since outcomes vary, no universal eventual monotonicity exists for the sum. This highlights that eventual monotonicity is not closed under addition.

📖 Explanation: Option A: Signs alternate forever. Values jump between positive and negative. Never monotone. Option B: Derivative $1 + \cos x \geq 0$. Non-decreasing everywhere. Option C: Related to Stirling; eventually increasing/decreasing? Actually $\ln(n!) \approx n \ln n - n$. Difference approaches constant. Detailed analysis shows it's monotone. Option D: $\arctan n$ increases to $\pi/2$. Perturbation $(-1)^n/n$ decays. For large $n$, derivative of arctan ($1/(1+x^2)$) dominates perturbation derivative. Eventually monotone. Only A retains perpetual sign alternation preventing any monotonic trend. Visualizing A shows zigzag crossing zero forever.

📖 Explanation: In many analytical proofs, establishing that $\lim r_n = L < 1$ suffices to claim eventual strict decrease. The epsilon-delta definition guarantees some $N$ exists where the condition holds. Explicitly finding $N$ is often algebraically tedious and unnecessary for proving existence/convergence. The logical structure relies on the existential quantifier in the limit definition. This distinguishes constructive algorithms (needing explicit $N$) from existence proofs (relying on guaranteed $N$). Understanding this abstraction is key to higher-level analysis.

📖 Explanation: Monotonicity of recursive sequences $x_{n+1}=f(x_n)$ depends on whether $f(x) > x$ (increasing) or $f(x) < x$ (decreasing) in the region where the sequence lives. Solving $(x^2+3)/4 > x \Rightarrow x^2 - 4x + 3 > 0 \Rightarrow (x-1)(x-3) > 0$. This holds for $x < 1$ or $x > 3$. Since $x_1=0.5$ and fixed point is 1, the sequence stays in $[0.5, 1)$. In this interval, $f(x) > x$, guaranteeing strict increase. Verifying this functional inequality on the invariant interval is the rigorous step linking recursion to eventual monotonicity.