📝 Divergence test for series (33 MCQs)

📖 Explanation: The Divergence Test states that if $\lim_{k \to \infty} u_k \neq 0$, the series $\sum u_k$ must diverge. The student correctly found the limit to be 1, but incorrectly interpreted this as the sum of the series. In reality, if the terms do not approach zero, the partial sums cannot settle to a finite value. The limit of the terms describes the behavior of individual addends, not the accumulated total. This distinction between sequence convergence and series convergence is critical.

📖 Explanation: The Divergence Test requires checking if $\lim_{k \to \infty} u_k = 0$. For $u_k = \sin(k)$, the sequence oscillates indefinitely between -1 and 1 and never settles at a single value, let alone zero. Because the limit does not exist (and specifically does not equal zero), the necessary condition for convergence fails. Consequently, the series diverges. This application highlights that 'limit equals zero' includes the requirement that the limit actually exists.

📖 Explanation: Visual inspection of the sequence's graph reveals a horizontal asymptote at $y = 0.001$. Mathematically, this means $\lim_{k \to \infty} a_k = 0.001$. According to the Divergence Test, since this limit is strictly non-zero, the infinite sum of these terms must grow without bound. Even though the terms are small, adding a constant non-zero value infinitely many times results in divergence. Graphs can sometimes deceive students into thinking 'close to zero' means 'convergent,' but the threshold is exactly zero.

📖 Explanation: This represents the most common misconception regarding the Divergence Test. The theorem states: If $\sum u_k$ converges, then $\lim u_k = 0$. The converse is false. While $\lim_{k \to \infty} \frac{1}{k} = 0$ satisfies the necessary condition, it does not guarantee convergence. The harmonic series is the classic counterexample where terms vanish yet the sum diverges. Students must distinguish between the test proving divergence (non-zero limit) and failing to prove convergence (zero limit).

📖 Explanation: The Divergence Test and the Integral Test are distinct tools. The Divergence Test checks only if $\lim_{k \to \infty} f(k) = 0$. It is possible for an integral to diverge while the sequence terms still approach zero (e.g., $f(x)=1/x$). Conversely, if the integral diverges because the function doesn't decay, the Divergence Test would also show divergence. However, knowing *only* that the integral diverges does not automatically tell us the limit of the terms; we must evaluate the limit separately to use the Divergence Test.

📖 Explanation: To apply the Divergence Test, we examine $\lim_{k \to \infty} a_k$. Since $a_{k+1} - a_k = 1/k^2 > 0$, the sequence is strictly increasing. Starting at $a_1=1$, every subsequent term is larger than 1. Therefore, $\lim_{k \to \infty} a_k \geq 1 \neq 0$. Because the terms themselves do not approach zero (they actually grow), the series $\sum a_k$ must diverge. This problem tests understanding of sequence behavior derived from recursion rather than explicit formulas.

📖 Explanation: The Alternating Series Test requires two conditions: decreasing magnitude and a limit of zero. Here, $\lim_{n \to \infty} (1 - e^{-n}) = 1 - 0 = 1$. Since the magnitude of the terms approaches 1 rather than 0, the fundamental prerequisite for any convergent series is violated. The Divergence Test immediately identifies this failure. The oscillation between values near +1 and -1 prevents the partial sums from settling, confirming divergence regardless of the alternating sign.

📖 Explanation: The original series has terms approaching 1/2, so it diverges. Options A, B, and D result in limits of 1, 0 (wait, 1/2 - 0.5 = 0? No, limit is 0 only if we divide), or 1.5 respectively. Actually, subtracting 0.5 gives limit 0. Let's re-evaluate. Original limit is 1/2. Dividing by $k$ gives $\frac{1}{2k+1}$, whose limit is 0. When the limit is 0, the Divergence Test provides no information about convergence or divergence; it is inconclusive. This distinguishes 'proving divergence' from 'failing to prove divergence.'

📖 Explanation: This is the contrapositive formulation of the Divergence Test. The theorem states: If a series converges, then its terms must approach zero. This is a necessary condition. Note that it does not imply monotonicity (terms can oscillate while decaying), absolute convergence (conditional convergence exists), or positivity. Understanding this one-way implication is foundational to analyzing infinite series correctly.

📖 Explanation: The student correctly notes the argument $1/k \to 0$. However, continuity of cosine implies $\lim_{k \to \infty} \cos(1/k) = \cos(0) = 1$. The Divergence Test looks at the limit of the *terms*, not their arguments. Since the terms approach 1, the series diverges. This error arises from conflating input behavior with output behavior in composite functions, a critical distinction in calculus analysis.

📖 Explanation: For series A, $\lim \frac{n}{n+1} = 1 \neq 0$, so the Divergence Test successfully proves divergence. For series B, $\lim \frac{n}{n^2+1} = 0$. When the limit is zero, the Divergence Test yields no conclusion; one must resort to other methods like the Limit Comparison Test or Integral Test. This comparison illustrates the test's asymmetric power: it is a definitive filter for divergence but useless for confirming convergence.

📖 Explanation: Numerical evidence can be misleading when terms decay extremely slowly or are very small but non-zero. If $\lim u_k = 10^{-9} \neq 0$, the Divergence Test guarantees divergence. The partial sums will eventually grow linearly with slope $10^{-9}$, but this growth is imperceptible until $n \approx 10^9$. This scenario emphasizes that analytical proofs supersede numerical observation, especially when dealing with limits that are non-zero but computationally negligible.

📖 Explanation: The Divergence Test applies to *any* series, alternating or not. We check $\lim_{k \to \infty} u_k$. Here, the sequence $\frac{(-1)^k k}{k+1}$ oscillates between values approaching +1 and -1. Since the limit does not exist (and certainly isn't zero), the series diverges. Students often mistakenly believe the Divergence Test requires positive terms or that alternating signs automatically suggest conditional convergence. The test simply asks: do the terms vanish?

📖 Explanation: Simplifying $\frac{k!}{(k+1)!}$ yields $\frac{1}{k+1}$. Recognizing this as the harmonic series (shifted) immediately identifies it as a canonical divergent series. While the student correctly found the limit is 0, recognizing the algebraic structure provides deeper insight than just applying a test. This connects the Divergence Test's inconclusive result to known series behaviors, reinforcing that 'limit zero' encompasses both convergent and divergent cases.

📖 Explanation: Scalar multiplication preserves the non-zero nature of the limit. If $\lim a_k = L \neq 0$, then $\lim (c \cdot a_k) = cL$. Since $c \neq 0$ and $L \neq 0$, the product $cL \neq 0$. Thus, the scaled series also fails the nth-term test and must diverge. This property reflects the linearity of limits and reinforces that divergence due to non-vanishing terms is robust under non-zero scaling.

📖 Explanation: Rationalizing the numerator gives $\frac{1}{\sqrt{k+1}+\sqrt{k}}$, which clearly approaches 0. Thus, the Divergence Test is inconclusive. However, evaluating the partial sum $S_n = \sqrt{n+1} - \sqrt{1}$ shows $S_n \to \infty$. This exemplifies a series where terms vanish (passing the necessary condition) yet the sum diverges. It highlights why additional tests are needed when the Divergence Test returns 'inconclusive,' particularly for telescoping structures.

📖 Explanation: If partial sums $S_n$ oscillate with constant amplitude, the difference $u_n = S_n - S_{n-1}$ does not approach zero; instead, it continues to jump back and forth. Specifically, $|u_n|$ remains significant enough to sustain the oscillation. Therefore, $\lim u_n \neq 0$, and the Divergence Test confirms divergence. Interpreting partial sum graphs to infer term behavior is a higher-order skill linking sequence and series concepts visually.

📖 Explanation: In the Divergence Test, $\lim u_k = 0$ is merely a prerequisite; its absence proves divergence, but its presence proves nothing. In the Alternating Series Test, $\lim b_k = 0$ combined with monotonic decrease is *sufficient* to guarantee convergence. Confusing these roles leads to errors like assuming 'limit zero' always means convergence. Understanding the logical status (necessary vs. sufficient) of this condition across different tests is crucial.

📖 Explanation: For rational functions, the limit at infinity is determined entirely by the leading terms. Since $\lim \frac{k^2}{2k^2} = 1/2 \neq 0$, the precise value of lower-order terms is irrelevant for the Divergence Test. The key is establishing non-zeroness, not exactness. This validates efficient asymptotic analysis as a legitimate tool for applying the Divergence Test, saving unnecessary computation while maintaining rigor.

📖 Explanation: Once $\lim u_k = 0$ is established, the Divergence Test has exhausted its utility. Re-applying it is futile. The appropriate next step depends on the series' structure (e.g., Ratio for factorials, Integral for integrable functions). This question targets procedural metacognition: knowing when a tool has finished its job. Students sometimes loop on the Divergence Test hoping for convergence proof, which it can never provide.

📖 Explanation: Mathematically, when $r=1$, the term becomes $1^n(0) = 0$? Wait, if $r=1$, $1-r=0$, so terms are 0. But usually geometric models fail at boundary. Let's assume the prompt implies the standard geometric form $\sum r^n$ or similar where limit isn't zero. If the intended series was $\sum 1$, terms don't vanish. Physically, non-vanishing terms in a cumulative sum mean the quantity increases/decreases forever without bound. In population models, this signals model invalidity or unchecked growth, translating mathematical divergence to real-world impossibility.

📖 Explanation: Despite the piecewise definition and lack of monotonicity, both subsequences $1/k$ and $1/k^2$ approach 0. Therefore, the unified sequence $a_k \to 0$. The Divergence Test is thus inconclusive. This dispels the myth that irregular or non-monotone sequences automatically fail the nth-term test. As long as all paths lead to zero, the necessary condition is met, requiring more sophisticated tools to determine actual convergence.

📖 Explanation: Options A, B, and D have terms with non-zero limits or non-existent limits, making them immediate candidates for the Divergence Test. Option C has terms $\frac{1}{n \ln n} \to 0$. Since the limit is zero, the Divergence Test cannot establish divergence (even though the series actually diverges by the Integral Test). Identifying series where the Divergence Test is powerless is as important as identifying where it works.

📖 Explanation: The student attributes the wrong theorem to the result. The Divergence Test says non-zero limit => divergence." It does NOT say "zero limit => possible divergence." That latter insight comes from studying specific series like the harmonic series via Integral or Cauchy Condensation tests. Conflating the test's statement with general series theory muddles logical foundations. Precision in attributing mathematical truths to their correct sources is vital."

📖 Explanation: This encapsulates the inconclusive nature of the zero-limit case. Examples include $\sum 1/k^2$ (convergent) and $\sum 1/k$ (divergent). Both satisfy $\lim u_k = 0$. Therefore, this condition partitions the universe of series into 'definitely divergent' (non-zero limit) and 'undetermined' (zero limit). Recognizing this partition prevents premature conclusions and guides selection of subsequent analytical tools.

📖 Explanation: Recognizing the definition of Euler's number $e$ allows immediate application of the Divergence Test. Since $e \approx 2.718 \neq 0$, the series diverges instantly. Using Ratio or Root tests would involve unnecessary complexity. This rewards pattern recognition and knowledge of fundamental limits over rote algorithmic application. Efficiency in problem-solving often comes from matching series forms to their most direct theoretical consequences.

📖 Explanation: While Stirling's approximation shows $k! \sim \sqrt{2\pi k}(k/e)^k$, the denominator $k^k$ grows much faster since $(k/e)^k / k^k = e^{-k} \to 0$. Thus, terms vanishing is consistent with asymptotic analysis. The Divergence Test confirming limit 0 aligns with growth hierarchy knowledge. This connects discrete series behavior to continuous growth rate intuition, validating that 'limit zero' is the expected outcome for super-exponential denominators.

📖 Explanation: Visual identification of a non-zero horizontal asymptote directly translates to $\lim u_k \neq 0$. This visual cue triggers the Divergence Test conclusion without calculation. Options A, B, and D are consistent with convergence (though not proof). Training visual literacy to spot 'non-zero floors' in sequence plots enables rapid triage of series problems before engaging algebraic machinery.

📖 Explanation: Adding 1 shifts every term by 1. If $\lim a_k = 0$, then $\lim (a_k + 1) = 1 \neq 0$. By Divergence Test, this new series must diverge. Other options preserve the zero limit (adding convergent terms, scaling, alternating signs). This tests understanding of how algebraic operations affect the necessary condition for convergence. Adding a non-vanishing constant is the simplest way to break convergence via the nth term.

📖 Explanation: Finding limit 0 is merely the entry gate, not the destination. For $\frac{\ln k}{k}$, despite vanishing terms, the series diverges (Integral Test). The student's error was treating the Divergence Test's inconclusive result as conclusive. The omitted step was selecting an appropriate follow-up test. This highlights workflow discipline: 'limit zero' triggers search for stronger tools, not cessation of analysis.

📖 Explanation: Convergence of series in reals is tied to Cauchy sequences. If $u_k \not\to 0$, then for any N, there exist m,n > N with |S_m - S_n| >= |u_k| > epsilon. Thus partial sums aren't Cauchy. Completeness equates Cauchy with convergent. So non-Cauchy => divergent. This deep connection shows the Divergence Test isn't just a heuristic but rooted in the foundational axioms of real analysis, linking basic calculus to advanced theory.

📖 Explanation: Both subsequences approach 0, so overall limit is 0. Divergence Test inconclusive. However, primes are sparse enough that $\sum_{p} 1/p$ diverges very slowly, but combined with $\sum 1/k^3$, the whole series actually diverges because sum of reciprocals of primes diverges. But Divergence Test can't see this. This exotic example reinforces that 'limit zero' hides complex behaviors dependent on number-theoretic properties, far beyond the scope of the nth-term test.

📖 Explanation: At t=0, $e^0 = 1$, so terms are $\sin(k)$. As established, $\sin(k)$ has no limit. Physically, this corresponds to undamped oscillation where energy doesn't dissipate, leading to non-convergent superposition. The Divergence Test mathematically flags this physical singularity. Connecting test outcomes to system stability demonstrates applied mathematics competency beyond pure computation.