📝 Radius and interval of convergence (37 MCQs)

📖 Explanation: According to the fundamental theorem of power series, if a series converges at a specific value $x = x_1$, it converges absolutely for all $x$ such that $|x| < |x_1|$. Since $|-3| = 3$ and $3 < 4$, the point $x = -3$ is strictly interior to the disk of convergence determined by $x=4$. Therefore, absolute convergence is guaranteed at $x=-3$ irrespective of whether the convergence at $x=4$ was conditional or absolute. This tests understanding of the geometric nature of convergence intervals centered at the origin.

📖 Explanation: The ratio test determines the open interval of absolute convergence but provides no information about the boundary points where the limit of the ratio equals exactly 1. For this series, at $x=3$, the series becomes the harmonic series $\sum 1/k$ which diverges, while at $x=1$, it becomes the alternating harmonic series $\sum (-1)^k/k$ which converges conditionally. Thus, the correct interval is $[1, 3)$. Failing to test endpoints individually is a common procedural error that leads to incomplete solutions in power series problems.

📖 Explanation: For a real power series centered at $x_0$, the radius of convergence $R$ is the distance from the center to the nearest singularity in the complex plane. On the real line, vertical asymptotes represent singularities where the function ceases to be analytic. Since the nearest singularities are at $\pm 5$ and the center is 0, the radius of convergence cannot exceed 5. Furthermore, since the function is smooth between them, the series converges up to these points, making $R=5$. This connects graphical analysis of singularities directly to the analytic concept of convergence radius.

📖 Explanation: Generally, the radius of convergence of a sum is at least the minimum of the individual radii. However, if the coefficients $a_k$ and $b_k$ are related such that their sum decays faster than either sequence individually (e.g., $b_k = -a_k$ for large $k$), the resulting series may have a larger radius of convergence. For instance, if $a_k = 1/3^k$ and $b_k = -1/3^k + 1/10^k$, then $R_a=3, R_b=3$, but $a_k+b_k = 1/10^k$, yielding $R=10$. Thus, assuming $R_S = \min(R_a, R_b)$ ignores potential cancellation effects.

📖 Explanation: The Ratio Test is specifically designed for series involving factorials and exponentials because the ratio $\frac{u_{k+1}}{u_k}$ allows for algebraic cancellation of factorial terms (e.g., $(2(k+1))! / (2k)!$). The Root Test would require evaluating $\sqrt[k]{k!}$, which is more complex and often requires Stirling's approximation. The Integral Test applies to functions, not discrete factorial sequences directly. Therefore, recognizing the structural cue of factorials immediately points to the Ratio Test as the optimal tool for determining the radius of convergence efficiently.

📖 Explanation: In applied modeling, the radius of convergence corresponds to the distance to the nearest singularity or phase transition in the physical system. Since the model works at $\Delta T = 100$, the radius $R$ must be $\geq 100$. Since it fails at $\Delta T = 150$, $R$ must be $\leq 150$. It is unlikely to be exactly 150 unless the failure point coincides precisely with a mathematical singularity, but physically, the valid domain is bounded by these observations. This question links abstract convergence concepts to empirical validation in scientific modeling.

📖 Explanation: The Ratio Test states that if $\lim |u_{k+1}/u_k| = L < 1$, the series converges absolutely, and if $L > 1$, it diverges. However, if $L = 1$, the test yields absolutely no information. In this specific case, at $x=1$, the series becomes $\sum 1/k^2$, which is a convergent p-series. The student erroneously treated the inconclusive case $L=1$ as a divergence criterion. Endpoints must always be tested using separate methods like the p-series test, integral test, or alternating series test.

📖 Explanation: Substituting $x^2$ for $x$ transforms the series into $\sum c_k (x^2)^k = \sum c_k x^{2k}$. For this new series to converge, we require $|x^2| < R$, which simplifies to $|x| < \sqrt{R}$. Thus, the radius of convergence becomes $\sqrt{R}$. Differentiation and integration preserve the original radius $R$. Multiplying by $x$ shifts indices but does not change the radius. This tests understanding of how variable substitutions affect the domain of convergence versus operations that preserve it.

📖 Explanation: The Cauchy product of two power series converges absolutely at least on the intersection of their intervals of absolute convergence. Series A converges absolutely on $(-2, 2)$. Series B converges absolutely on $(-3, 3)$ (note: convergence at endpoint 3 may be conditional). The intersection of the open intervals of absolute convergence is $(-2, 2)$. While the product might converge on a larger set depending on specific coefficients, it is only mathematically guaranteed on the smaller of the two radii. This emphasizes the distinction between guaranteed bounds and potential extensions.

📖 Explanation: Rewriting the series as $\sum \frac{2^k}{k!} x^k$, applying the ratio test yields $\lim |\frac{2^{k+1} x^{k+1}}{(k+1)!} \cdot \frac{k!}{2^k x^k}| = \lim |\frac{2x}{k+1}| = 0$ for any fixed $x$. Since the limit is 0 for all real numbers, the series converges everywhere, meaning $R = \infty$. The first student likely confused this with a geometric series $\sum (2x)^k$ where $R=1/2$. Recognizing the dominance of factorials over exponentials is crucial for correctly identifying infinite radii in Taylor-type series.

📖 Explanation: Differentiation of a power series preserves the radius of convergence $R$, so $g(x)$ also has $R=4$ centered at 3, giving the open interval $(-1, 7)$. However, differentiation can destroy convergence at the endpoints. A series might converge conditionally at an endpoint while its derivative diverges there (e.g., $\sum x^k/k^2$ vs $\sum x^{k-1}/k$). Therefore, while the interior is safe, the endpoints of the derived series must be tested independently and cannot be assumed to match the original series' endpoint behavior.

📖 Explanation: This series lacks odd powers of $(x-5)$, making it a power series in the variable $u = (x-5)^2$. Applying the ratio test to $u$ gives convergence for $|u| < 4$. Translating back to $x$, we get $|x-5|^2 < 4 \implies |x-5| < 2$. While the numerical answer $R=2$ is correct for $x$, conceptualizing it purely as a standard series in $(x-5)$ obscures the structure. Standard formulas for radius $R = \lim |c_k/c_{k+1}|$ assume consecutive powers; here, coefficients for odd powers are zero, causing standard ratio limits on $c_k$ to fail or oscillate. One must treat it as a series in $(x-5)^2$.

📖 Explanation: For $\sum k! x^k$, the ratio test gives $\lim |(k+1)! x^{k+1} / k! x^k| = \lim |(k+1)x| = \infty$ for any $x \neq 0$. Thus, the series diverges for all non-zero $x$ and converges only at the center $x=0$. This represents the extreme case where $R=0$. Geometric series with $r=1$ diverge everywhere except trivially, and $e^x$ converges everywhere. Alternating decreasing series typically have $R \geq 1$. Understanding $R=0$ cases is vital for recognizing the limitations of power series representations.

📖 Explanation: In mathematical modeling, the radius of convergence defines where the series sums to a finite value, but it does not guarantee the model's fidelity to reality. Physical systems often have domains of validity narrower than the mathematical convergence interval due to unmodeled factors (like saturation). The correct interpretation distinguishes between analytical convergence ($R=10$) and applicability ($t \leq 8$). Blindly trusting the mathematical radius without considering domain-specific constraints is a common pitfall in applied calculus and engineering.

📖 Explanation: Integration increases the denominator by a factor of roughly $k$, which can turn a divergent endpoint into a convergent one. For example, $\sum x^k$ diverges at $x=1$, but its integral $\sum x^{k+1}/(k+1)$ converges at $x=1$ (alternating harmonic-like behavior after substitution/sign adjustment). Conversely, if divergence is strong enough, integration might not fix it. Thus, unlike the interior where behavior is identical, endpoints require independent testing after integration. This highlights the asymmetry between differentiation (which worsens endpoint convergence) and integration (which can improve it).

📖 Explanation: While the numerical radius happens to be 1 in this specific case (since $\sqrt{1}=1$), the student's reasoning 'coefficients are the same' is dangerous. If the original radius were 4, the new radius would be $\sqrt{4}=2$, not 4. The presence of $x^{2k}$ means the series behaves like $\sum y^k/k$ with $y=x^2$. The convergence condition is on $y$, which translates to a square root constraint on $x$. Correct reasoning must account for the power transformation, not just coefficient similarity, to avoid errors in general cases.

📖 Explanation: Power series convergence is determined by the distance to the nearest singularity in the complex plane. For $\ln(x)$ centered at 1, the function has a branch point/singularity at $x=0$. The distance from center 1 to singularity 0 is exactly 1. Even though $\ln(x)$ is smooth for $x>2$, the series cannot converge past the barrier imposed by the singularity at 0. This illustrates that real-domain definition does not dictate series radius; complex-analytic structure does. Students often confuse function domain with series convergence interval.

📖 Explanation: Recognizing series types saves time. $\sum x^k/k!$ is the Maclaurin series for $e^x$, known to have $R=\infty$; even without memorization, the ratio test quickly yields limit 0. $\sum (x/2)^k$ is geometric with ratio $x/2$, converging for $|x/2|<1 \implies |x|<2$. Identifying these structures avoids unnecessary mechanical computation. Efficiency in HOTS problems comes from pattern recognition and selecting the appropriate mental model rather than blindly applying algorithms. This question assesses strategic problem-solving over rote procedure.

📖 Explanation: If all coefficients $c_k$ were non-negative, then $\sum c_k R^k$ and $\sum c_k (-R)^k$ would involve terms with magnitudes $|c_k R^k|$. Absolute convergence at $x=a+R$ would imply $\sum |c_k R^k| < \infty$, which would force absolute (and thus conditional) convergence at $x=a-R$ as well. Since convergence behaviors differ at the two endpoints, the series cannot consist solely of positive terms; there must be sign variations or cancellations allowing conditional convergence at one end but not the other. This probes deep understanding of absolute vs. conditional convergence symmetry.

📖 Explanation: Even inside the interval of convergence, the rate of convergence is not uniform. Near the boundary $R$, the terms $c_k x^k$ decay much more slowly than near the center. Consequently, a fixed number of terms provides high accuracy centrally but poor accuracy near the edge. This is a fundamental property of power series approximation. Practical application requires adaptive truncation or alternative expansions near boundaries. Understanding this prevents misuse of series approximations in numerical computing where precision matters.

📖 Explanation: When $\lim |a_{k+1}/a_k|$ fails to exist, the Ratio Test is inapplicable. The Root Test uses $\limsup \sqrt[k]{|a_k|}$, which handles oscillating coefficients robustly. Here, $\sqrt[k]{2+(-1)^k} \to 1$, giving $R=1$. Alternatively, decomposing into $2\sum x^k + \sum (-x)^k$ reveals two geometric series both with $R=1$; their sum shares this radius. This tests adaptability when standard tools fail and knowledge of alternative methods like limsup or series decomposition. Rigid adherence to the ratio test is a common limitation in student problem-solving.

📖 Explanation: Integration improves convergence properties. At $x=4$, the original series is harmonic ($\sum 1/k$, divergent). The integrated series has terms $\sim 1/k^2$, which converges absolutely. Thus, the new interval likely includes $x=4$. At $x=2$, original is alternating harmonic (convergent); integrated is absolutely convergent. So the interval expands to $[2, 4]$. Assuming endpoint behavior is invariant under integration is a critical error. Students must understand that calculus operations on series can alter boundary inclusion even when preserving the radius.

📖 Explanation: Taylor series convergence is obstructed by singularities. Vertical asymptotes, cusps, or points of non-differentiability act as barriers. Horizontal asymptotes describe behavior at infinity and do not limit the local radius (e.g., $\arctan x$ has horizontal asymptotes but finite $R$ due to complex singularities, though visually on real line it looks smooth; however, vertical asymptotes are definitive real-line blockers). Local extrema and inflection points are features of smooth functions and do not impede convergence. Identifying visual singularities is key to predicting convergence domains from graphs.

📖 Explanation: Let $y = x^3$. The series becomes $\sum c_k y^k$, which converges for $|y| < 4$. Substituting back, $|x^3| < 4 \implies |x| < \sqrt[3]{4}$. This transformation compresses the interval of convergence because higher powers of $x$ grow faster, reaching the divergence threshold sooner. Students often mistakenly keep the original interval or apply square roots instead of cube roots. Correctly mapping variable substitutions to domain transformations is essential for handling composite power series.

📖 Explanation: The Ratio Test provides a limit $L$. If $L < 1$, absolute convergence; if $L > 1$, divergence. But at the boundary where $L=1$, the test is mathematically inconclusive. Since the interval of convergence includes potential endpoint convergence (conditional or absolute), the Ratio Test alone cannot fully specify the interval. Endpoint analysis requires supplementary tests. This is a foundational definition recall question ensuring students understand the scope and limitations of their primary tool.

📖 Explanation: Multiplying by $(x-c)^m$ simply shifts the indices of the series: $\sum a_k (x-c)^{k+m} = \sum b_j (x-c)^j$ where $b_j = a_{j-m}$ for $j \geq m$. The asymptotic behavior of the coefficients $b_j$ is identical to $a_k$ for large indices. Since radius depends on the tail behavior of coefficients, it remains unchanged. This tests understanding that radius is an asymptotic property invariant under finite shifts or polynomial multiplication, distinguishing it from operations like substitution that alter growth rates.

📖 Explanation: Coefficients are $c_n(t) = A_n e^{-\lambda_n t}$. As $t$ increases, $e^{-\lambda_n t}$ decays faster for larger $n$ (since $\lambda_n$ increases). This rapid suppression of high-order terms improves convergence properties, effectively increasing the radius. Physically, diffusion smooths out gradients, making the solution more analytic over larger spatial domains as transients die out. This connects PDE behavior to series convergence, illustrating interdisciplinary application of radius concepts.

📖 Explanation: The radius is tied to the argument of the series. $\sum c_k u^k$ converges for $|u|<5$. With $u=2x$, we need $|2x|<5$. The student likely viewed $R$ as an intrinsic constant attached to the coefficient sequence $c_k$ independent of the variable form. This static view of radius ignores the functional composition. Correct thinking treats radius as a constraint on the entire term $c_k(\text{variable})^k$. Identifying this specific misconception helps target instruction on variable substitution.

📖 Explanation: By definition, a real power series $\sum c_k (x-a)^k$ converges on an interval $(a-R, a+R)$ possibly including endpoints. This interval is inherently symmetric around $a$. Asymmetry can only arise in endpoint inclusion (e.g., $[a-R, a+R)$), but the open interval of absolute convergence is always symmetric. Options B, C, D describe series that still possess symmetric open intervals. This question tests the fundamental geometric definition of power series domains, dispelling notions of inherent asymmetry in convergence regions.

📖 Explanation: Differentiation preserves the radius of convergence $R$. However, each differentiation can potentially cause divergence at endpoints where the original series converged conditionally. After two derivatives, the risk of losing endpoint convergence is compounded. For example, $\sum x^k/k^3$ converges at $x=1$; first derivative $\sum x^{k-1}/k^2$ converges; second derivative $\sum x^{k-2}/k$ diverges at $x=1$. Thus, while $R$ is stable, the closed interval of convergence can shrink. This tests nuanced understanding of operational effects on series domains.

📖 Explanation: These are canonical examples representing extremes. $1/k!$ decays super-exponentially, allowing convergence for all $x$ ($R=\infty$). $k!$ grows super-exponentially, overwhelming any $x^k$ for $x \neq 0$ ($R=0$). This contrast establishes the direct inverse relationship between coefficient growth rate and convergence radius. Understanding these benchmarks allows quick estimation of radii for intermediate cases. It reinforces that radius is fundamentally a measure of coefficient asymptotics.

📖 Explanation: The Root Test states convergence when $\lim \sqrt[k]{|a_k x^k|} = L|x| < 1$. Here $L = 1/e$. So $|x|/e < 1 \implies |x| < e$. The radius is the reciprocal of the limit $L$. The student correctly computed the limit but failed to take the reciprocal to find $R$. Confusing the limit value with the radius itself is a frequent algebraic slip. This question targets precise application of the convergence criterion formula.

📖 Explanation: Near $R$, terms decay slowly. Achieving machine precision might require millions of terms, introducing round-off errors and computation time. Acceleration techniques like Euler summation transform slowly convergent series into rapidly convergent ones without changing the limit. This bridges pure analysis (existence of sum) and numerical analysis (efficient computation). Recognizing that theoretical convergence ≠ practical computability is a higher-order insight for applied mathematics.

📖 Explanation: The product of two analytic functions is analytic wherever both are analytic. Both series converge absolutely for $|x| < 3$. Thus, their Cauchy product converges absolutely for $|x| < 3$. While $h(x)$ might be analytically continuable beyond $|x|=3$ (if singularities of $f$ are removable in the product), we are only guaranteed convergence on the intersection of the original disks. Hence, $R_h \geq 3$. This tests conservative reasoning about combined domains versus optimistic assumptions.

📖 Explanation: This is the classic counterexample showing real smoothness ≠ analyticity globally. The function has poles at $\pm i$ in the complex plane. The radius of convergence of a real Taylor series is the distance to the nearest complex singularity. Distance from 0 to $\pm i$ is 1. Thus $R=1$. Real-variable intuition fails here. Understanding this requires bridging real and complex analysis, a pinnacle concept in series theory explaining why some smooth functions have finite series domains.

📖 Explanation: The student confuses the transformation rule. Let $u=x^2$. Original series converges for $|u|<R$. Substitute back: $|x^2|<R \implies |x|<\sqrt{R}$. The radius transforms via the inverse of the power substitution. Claiming $R^2$ suggests misunderstanding functional composition. This error arises from misapplying algebraic rules to analytic domains. Correcting it reinforces the link between variable change and domain scaling.