📝 Ratio test for convergence (37 MCQs)

📖 Explanation: To apply the Ratio Test to $\sum \frac{k!}{k^k}$, we examine $\frac{(k+1)!}{(k+1)^{k+1}} \cdot \frac{k^k}{k!}$. This simplifies to $\frac{k+1}{1} \cdot \frac{k^k}{(k+1)^{k+1}} = \frac{k^k}{(k+1)^k} = \left( \frac{k}{k+1} \right)^k = \frac{1}{(1+1/k)^k}$. As $k \to \infty$, the denominator approaches $e$, so the limit is $1/e$. Since $1/e < 1$, the series converges absolutely. The misconception that factorials and exponentials grow similarly ignores the specific base structure in the denominator.

📖 Explanation: When the limit of the ratio $a_{k+1}/a_k$ exists and equals $L < 1$, the series behaves asymptotically like a geometric series with common ratio $L$. Specifically, for any $\epsilon > 0$, there exists an $N$ such that for $k > N$, $a_{k+1} < (L+\epsilon)a_k$. This implies the tail of the series is bounded by a convergent geometric series with ratio $L+\epsilon$. Thus, the terms exhibit exponential decay characteristic of geometric sequences, distinguishing them from p-series which decay polynomially.

📖 Explanation: The Ratio Test states that if $\lim |a_{k+1}/a_k| = L < 1$, the series converges; if $L > 1$, it diverges. However, if $L = 1$, the test provides no information. For rational functions like $\frac{k^2+1}{k^3+k} \approx \frac{1}{k}$, the ratio of consecutive terms always approaches 1 regardless of whether the series converges or diverges. Therefore, obtaining L=1 necessitates switching to a different test, such as the Limit Comparison Test or Integral Test, rather than drawing a conclusion about divergence based solely on the Ratio Test result.

📖 Explanation: By applying the Ratio Test to the power series $\sum c_k (x-a)^k$, we evaluate $\lim \left| \frac{c_{k+1}(x-a)^{k+1}}{c_k(x-a)^k} \right| = |x-a| \lim \left| \frac{c_{k+1}}{c_k} \right| = L|x-a|$. For convergence, we require this limit to be strictly less than 1, i.e., $L|x-a| < 1 \implies |x-a| < 1/L$. By definition, the radius of convergence $R$ is the value such that the series converges for $|x-a| < R$. Comparing these inequalities shows directly that $R = 1/L$. This relationship is foundational for finding intervals of convergence efficiently.

📖 Explanation: The Ratio Test is specifically powerful for series containing factorials or products because the recursive nature of factorials ($(n+1)! = (n+1)n!$) allows for significant algebraic cancellation when forming the quotient $a_{k+1}/a_k$. In contrast, the Root Test would require evaluating $\sqrt[k]{(k!)^2/(2k)!}$, which involves Stirling's approximation or advanced limits. Direct comparison is also difficult due to the double factorial structure. Thus, the algebraic simplification afforded by division makes the Ratio Test the optimal choice for this specific structural form.

📖 Explanation: The Ratio Test relies on the existence of the limit $L = \lim |a_{k+1}/a_k|$. For this piecewise series, when $k$ is odd, $a_{k+1}/a_k = (1/3^{k+1}) / (1/2^k) = \frac{2^k}{3^{k+1}} \to 0$. When $k$ is even, $a_{k+1}/a_k = (1/2^{k+1}) / (1/3^k) = \frac{3^k}{2^{k+1}} \to \infty$. Since the subsequences of ratios converge to different limits (0 and infinity), the overall limit does not exist. Consequently, the standard Ratio Test is inapplicable, even though the series converges by comparison to $\sum (1/2)^k$. This highlights the necessity of the limit's existence for the test.

📖 Explanation: While the Ratio Test guarantees eventual convergence for $e^x$ everywhere, for large $x$ like 10, the terms $x^k/k!$ initially grow until $k \approx x$. During this growth phase, the ratio $a_{k+1}/a_k = x/(k+1) > 1$. A naive algorithm checking for ratio < epsilon might malfunction or misinterpret the increasing phase. Furthermore, even after terms start decreasing, the ratio being small doesn't guarantee the partial sum has reached desired precision relative to the true value, especially given floating-point accumulation errors. The test ensures convergence theoretically but doesn't directly bound truncation error for finite sums.

📖 Explanation: The original series $\sum \frac{k!}{k^k}$ has ratio limit $1/e < 1$, so it converges. If we replace the denominator $k^k$ with $k!$, the series becomes $\sum \frac{k!}{k!} = \sum 1$. The ratio of consecutive terms is exactly 1 for all $k$. Thus, $\lim |a_{k+1}/a_k| = 1$, rendering the Ratio Test inconclusive (though the series obviously diverges by the Divergence Test). Replacing numerator with $k^2$ still yields limit 0. Alternating signs don't affect the absolute ratio limit. Index shifting doesn't change the asymptotic limit. Only changing the growth rate to match factorials creates the L=1 boundary case.

📖 Explanation: The graph indicates that $\lim_{k \to \infty} r_k = 1$. According to the Ratio Test, when the limit of the ratio equals exactly 1, the test yields no information about convergence or divergence. The series could be a convergent p-series like $\sum 1/k^2$ (where ratio $\to 1$) or a divergent harmonic series $\sum 1/k$ (where ratio $\to 1$). Visual confirmation of the limit approaching 1 only confirms the test's failure, necessitating alternative methods like the Integral Test or Comparison Test to determine the series' actual behavior.

📖 Explanation: The correct ratio setup is $\frac{(2k+2)!}{4^{k+1}((k+1)!)^2} \cdot \frac{4^k (k!)^2}{(2k)!}$. Simplifying factorials: $\frac{(2k+2)(2k+1)}{4(k+1)^2}$. Expanding numerator: $4k^2 + 6k + 2$. Denominator: $4(k^2+2k+1) = 4k^2 + 8k + 4$. The limit as $k \to \infty$ is the ratio of leading coefficients: $4/4 = 1$. The student's error was claiming this rational function tends to infinity. Recognizing that highest-degree terms dominate reveals the limit is 1, making the Ratio Test inconclusive. This is a classic pitfall with central binomial coefficients.

📖 Explanation: The series has the general term $a_k = (b_k)^k$ where $b_k = \frac{k+1}{2k}$. The Root Test evaluates $\sqrt[k]{a_k} = b_k$, whose limit is simply $\lim \frac{k+1}{2k} = 1/2$. This is immediate. Using the Ratio Test would require evaluating $\frac{((k+2)/(2k+2))^{k+1}}{((k+1)/2k)^k}$, which involves complex limits of the form $(1+1/n)^n$ and algebraic manipulation. While both work, the Root Test exploits the structural form $(\dots)^k$ directly, bypassing messy ratio algebra. Choosing the right test based on term structure saves significant computational effort.

📖 Explanation: From the recurrence, $\left| \frac{a_{k+1}}{a_k} \right| = \frac{|k-1|}{(k+1)^2}$. Applying the Ratio Test for the power series radius: $\lim_{k \to \infty} \left| \frac{a_{k+1} x^{k+1}}{a_k x^k} \right| = |x| \lim_{k \to \infty} \frac{k-1}{(k+1)^2}$. The limit of the coefficient ratio is $\lim \frac{k}{k^2} = 0$. Thus, the overall limit is $0 \cdot |x| = 0$ for any finite $x$. Since $0 < 1$ for all real $x$, the series converges everywhere. The radius of convergence is infinite. Recurrence relations often allow direct Ratio Test application without needing closed-form expressions for coefficients.

📖 Explanation: The limit $L=0.9$ means that for sufficiently large $k$, the ratio is approximately 0.9. Formally, for any $\epsilon > 0$, eventually $|a_{k+1}/a_k| < 0.9 + \epsilon$. This guarantees the tail behaves like a geometric series with ratio slightly larger than 0.9. However, it does not guarantee the ratio is always ≤ 0.9, nor does it account for pre-asymptotic behavior. The series could have large initial terms or ratios exceeding 0.9 before settling down. Thus, while asymptotic rates match, claiming strictly 'faster' convergence globally is unjustified without additional bounds on the entire sequence.

📖 Explanation: For $\sum \frac{\ln k}{k^2}$, applying the Ratio Test gives $\lim \frac{\ln(k+1)}{(k+1)^2} \cdot \frac{k^2}{\ln k}$. Evaluating this requires L'Hôpital's rule or asymptotic expansions for logarithms, resulting in limit 1 (inconclusive anyway). Even if it weren't 1, the algebra is cumbersome. In contrast, the Limit Comparison Test with $\sum 1/k^2$ immediately resolves convergence. The other options involve factorials or exponentials where Ratio Test simplifications are clean and conclusive. Recognizing when the Ratio Test becomes computationally inefficient compared to simpler tests is a key strategic skill in series analysis.

📖 Explanation: First, simplify the ratio of coefficients: $\frac{a_{k+1}}{a_k} = \frac{k \binom{2k}{k}}{(k+1) \binom{2k+2}{k+1}} = \frac{k}{k+1} \cdot \frac{(2k)! (k+1)!^2}{(k!)^2 (2k+2)!}$. After algebraic simplification, this reduces to $\frac{k(k+1)}{(2k+1)(2k+2)} \to \frac{1}{4}$. Including $x$, the ratio limit is $|x|/4$. Convergence requires $|x|/4 < 1 \implies |x| < 4$. At endpoints $x=\pm 4$, further testing is needed, but the open interval is definitely (-4, 4). This problem combines combinatorial identities with the Ratio Test, requiring careful factorial manipulation typical of advanced calculus problems.

📖 Explanation: The rigorous statement of the Ratio Test uses the limit superior. If $\limsup |a_{k+1}/a_k| = L < 1$, then there exists some $r$ with $L < r < 1$ and an integer $N$ such that for all $k \geq N$, $|a_{k+1}/a_k| < r$. This implies the tail of the series is dominated by a convergent geometric series with ratio $r$. Therefore, the series converges absolutely even if the ordinary limit fails to exist. This extension makes the Ratio Test applicable to certain oscillating sequences where the standard limit formulation would fail.

📖 Explanation: Applying Ratio Test: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{(2n+2)!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!} |z| = \frac{(2n+2)(2n+1)}{(n+1)^2} |z|$. Simplifying: $\frac{2(n+1)(2n+1)}{(n+1)^2} |z| = \frac{2(2n+1)}{n+1} |z|$. As $n \to \infty$, this approaches $4|z|$. For convergence, we need $4|z| < 1 \implies |z| < 1/4$. The student's claim of radius 1 is incorrect. This series relates to the binomial expansion of $(1-4z)^{-1/2}$, confirming the singularity at $z=1/4$. Proper factorial simplification is crucial here.

📖 Explanation: While the Root Test can determine convergence whenever the Ratio Test can (and sometimes when it can't), applying it to factorials requires evaluating $\sqrt[n]{n!}$, which is non-trivial without advanced tools like Stirling's formula. Conversely, the Ratio Test leverages the recursive definition $(n+1)! = (n+1)n!$ to cancel terms instantly, reducing the problem to basic algebra. This computational convenience makes the Ratio Test the practical standard for factorial series, reserving the Root Test for cases where ratios are intractable or involve pure nth powers.

📖 Explanation: If the ratio is bounded by $r < 1$ for $k \geq N$, then $a_{N+2} \leq r a_{N+1}$, $a_{N+3} \leq r^2 a_{N+1}$, etc. The remainder $R_N = a_{N+1} + a_{N+2} + \dots \leq a_{N+1}(1 + r + r^2 + \dots)$. Summing the geometric series gives $R_N \leq \frac{a_{N+1}}{1-r}$. This provides a concrete error bound derived directly from Ratio Test conditions, linking the abstract convergence criterion to practical numerical estimation. Note it starts with $a_{N+1}$, not $a_N$, because the remainder excludes the Nth term.

📖 Explanation: The Ratio Test yields $\lim \frac{(k+1)^{k+1}}{(k+1)!} \frac{k!}{k^k} = \lim (1+1/k)^k = e > 1$, confirming divergence. The Divergence Test notes $a_k = k^k/k!$. Since $k^k$ dominates $k!$ (as seen by Ratio Test > 1), $a_k \to \infty \neq 0$, so it diverges. Both arguments are correct and consistent. The Ratio Test actually proves the growth rate that justifies the Divergence Test observation. In practice, once you compute the ratio for the Ratio Test, you've essentially done the work for the Divergence Test too. Both validate the conclusion robustly.

📖 Explanation: The ratio of consecutive terms involves $\frac{(a+k)(b+k)}{(c+k)(k+1)} z$. As $k \to \infty$, the numerator behaves like $k^2$ and denominator like $k^2$. The limit of the coefficient part is 1. Thus, the overall ratio limit is $|z| \cdot 1 = |z|$. Convergence requires $|z| < 1$. This universality (R=1) is a defining characteristic of standard hypergeometric series $_2F_1$. Regardless of specific parameter values (unless terminating), the quadratic growth in numerator and denominator cancels perfectly, leaving unit radius. This structural insight explains why so many special functions share the same disk of convergence.

📖 Explanation: The standard Ratio Test requires the limit of ratios to exist. For $\sum \frac{\sin k}{2^k}$, the ratio of sines oscillates wildly and has no limit. However, absolute convergence can be established via Comparison: $|\frac{\sin k}{2^k}| \leq \frac{1}{2^k}$. Since $\sum (1/2)^k$ converges geometrically, the original series converges absolutely. Attempting to force the Ratio Test here is a category error; recognizing when structural bounds supersede ratio limits is critical. Absolute convergence implies convergence, resolving the issue without needing the problematic sine ratio limit.

📖 Explanation: The recurrence explicitly gives the ratio $\frac{a_{k+1}}{a_k} = \frac{2k}{3k+1}$. Taking the limit as $k \to \infty$ yields $2/3$. Since $L = 2/3 < 1$, the Ratio Test definitively establishes absolute convergence. Recursive definitions often provide the ratio directly, making the Ratio Test the natural tool. The starting value affects the sum but not the convergence property (as long as it's finite/nonzero). This illustrates how recursive structures map perfectly to Ratio Test methodology.

📖 Explanation: A 'false negative' in testing usually means failing to detect a true effect. Here, if a series converges but the Ratio Test returns L=1 (inconclusive), the test failed to confirm convergence. This happens frequently with p-series like $\sum 1/k^2$. Students might mistakenly interpret 'inconclusive' as 'diverges' or give up, missing the true convergence. Options B and C are mathematically impossible (Ratio Test is sound for L≠1). Option D is a limitation but L=1 is the most common practical pitfall where convergence exists but the test is blind to it, requiring backup methods.

📖 Explanation: For $e^{-20}$, terms $|a_k| = 20^k/k!$ increase until $k \approx 20$, reaching values near $20^{20}/20! \approx 4 \times 10^7$. In fixed-precision arithmetic, intermediate terms may overflow or lose precision before the decay phase begins. Even if overflow is avoided, summing huge alternating terms introduces catastrophic cancellation error. The Ratio Test guarantees mathematical convergence but ignores numerical stability. Practical implementation requires scaling, log-space computation, or alternative algorithms for large arguments. This highlights the gap between analytical convergence and computational feasibility.

📖 Explanation: Since $L_b = 2 > 1$, $b_k$ grows exponentially and $b_k \not\to 0$. Since $L_a = 0.5 < 1$, $a_k \to 0$ rapidly. The sum $a_k + b_k$ behaves asymptotically like $b_k$ because exponential growth dominates exponential decay. Specifically, $\lim (a_k + b_k) \neq 0$. By the Divergence Test, $\sum (a_k + b_k)$ must diverge. The Ratio Test on the sum itself would also show limit 2. Adding a convergent series to a divergent one always yields divergence. This reinforces that convergence is determined by the 'worst-behaved' component.

📖 Explanation: The student commits two errors: misinterpreting L=1 as convergence and assuming proximity to geometric behavior implies convergence. For $\sum \frac{1}{k \ln k}$, the ratio indeed approaches 1, but the series diverges (by Integral Test: $\int \frac{dx}{x \ln x} = \ln(\ln x) \to \infty$). Geometric series require ratio strictly less than 1. At L=1, decay is sub-geometric (polynomial/logarithmic), which may or may not suffice for convergence. Correct refutation requires citing the inconclusiveness of L=1 and applying a sensitive test like Integral or Cauchy Condensation to reveal the slow divergence.

📖 Explanation: The Ratio Test yielding L=1 correctly flags that the series is on the convergence/divergence boundary. It doesn't 'fail'; it accurately reports insufficient discrimination power at this scale. The refined analysis showing $a_k \sim e/k$ confirms divergence via comparison to harmonic series. This demonstrates that L=1 is a signal to zoom in with finer tools (asymptotics, comparison), not a defect. Understanding L=1 as a transition zone rather than a dead end is crucial for advanced series analysis. The test did its job by identifying the critical case.

📖 Explanation: If $c_n \sim n!$, then $|c_{n+1}/c_n| \sim n+1 \to \infty$. The Ratio Test limit is $\infty \cdot |\lambda| = \infty$ for any $\lambda \neq 0$. Thus, the series diverges for all nonzero coupling. Physically, this means perturbation series in QFT/QM are typically asymptotic expansions: useful for approximation at low orders but fundamentally non-convergent. Recognizing factorial growth via Ratio Test alerts physicists to the asymptotic nature of their tools, preventing erroneous assumptions about exact solvability via infinite summation. This connects mathematical divergence to physical interpretation.

📖 Explanation: Denominator: $(k+1)! - k! = k!(k+1-1) = k \cdot k!$. Term becomes $\frac{k!}{k \cdot k!} = \frac{1}{k}$. This is the harmonic series. Applying Ratio Test to $1/k$: $\frac{k}{k+1} \to 1$. Test is inconclusive. Student's intuition about divergence is correct but reasoning via raw Ratio Test on unsimplified form is risky. Simplification reveals the true nature. This emphasizes algebraic simplification BEFORE testing. Blind application to complex fractions can obscure simple underlying structures like harmonic series.

📖 Explanation: Let $y = x^2$. The series becomes $\sum a_k y^k$ with radius $R_y = 3$. Convergence requires $|y| < 3 \implies |x^2| < 3 \implies |x| < \sqrt{3}$. Thus, the new radius is $\sqrt{3}$. Students often mistakenly multiply or square R directly. Understanding substitution effects on radius is crucial for manipulating power series. The Ratio Test on original gives $\lim |a_{k+1}/a_k| = 1/3$. For new series, ratio is $|a_{k+1} x^{2k+2} / a_k x^{2k}| = |a_{k+1}/a_k| x^2 \to x^2/3$. Require $x^2/3 < 1 \implies |x| < \sqrt{3}$. Consistent result.

📖 Explanation: The Ratio Test analyzes internal structure via consecutive term ratios, requiring no external reference. The Limit Comparison Test fundamentally depends on selecting an appropriate benchmark series $b_k$ with known behavior. This makes LCT dependent on user insight/experience to choose $b_k$, while Ratio Test is algorithmic. However, Ratio Test fails for L=1 cases where LCT often succeeds with proper benchmark. Understanding this trade-off guides test selection: use Ratio for factorials/exponentials (self-similar structure); use LCT for rational/algebraic terms where benchmarks are obvious.

📖 Explanation: When consecutive term ratios are near 1, terms decay very slowly, necessitating vast numbers of terms for convergence (slow convergence). Simultaneously, if terms alternate or vary in sign, summing many nearly-equal-magnitude terms causes catastrophic cancellation in finite precision. Even for positive series, slow decay means partial sums change incrementally, amplifying roundoff accumulation. Thus, ratio ≈ 1 signals both theoretical inefficiency and numerical fragility. Acceleration techniques (Euler, Shanks) or analytic continuation become necessary. Recognizing this dual challenge links abstract convergence rates to practical computational constraints.

📖 Explanation: Using Stirling: $n! \sim \sqrt{2\pi n}(n/e)^n$. Then $\frac{(2k)!}{(k!)^2 4^k} \sim \frac{\sqrt{4\pi k}(2k/e)^{2k}}{2\pi k (k/e)^{2k} 4^k} = \frac{2\sqrt{\pi k} 4^k k^{2k} e^{-2k}}{2\pi k k^{2k} e^{-2k} 4^k} = \frac{1}{\sqrt{\pi k}}$. Since $\sum 1/\sqrt{k}$ diverges (p=0.5), the original series diverges. Ratio Test L=1 missed this. This exemplifies using asymptotic analysis to resolve borderline Ratio Test cases, connecting combinatorics to analytic number theory techniques essential for advanced series evaluation.