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πŸ“ Alternating series test (37 MCQs)

πŸ“– From Calculus β€’ 10. Infinite Series in Calculus β€’ 37 questions available

What is Alternating series test?

If bn>0b_n > 0, bn+1≀bnb_{n+1} \le b_n (decreasing), and lim⁑nβ†’βˆžbn=0\lim_{n\to\infty} b_n = 0, then the alternating series βˆ‘n=1∞(βˆ’1)n+1bn\sum_{n=1}^{\infty} (-1)^{n+1} b_n converges; for example, βˆ‘(βˆ’1)n+11n\sum (-1)^{n+1} \frac{1}{n} converges (to ln⁑2\ln 2) even though the harmonic series diverges.

6
Easy
12
Medium
19
Hard

πŸ“ All Alternating series test MCQs

Q1. A student claims that the series βˆ‘(βˆ’1)nan\sum (-1)^n a_n converges solely because lim⁑nβ†’βˆžan=0\lim_{n \to \infty} a_n = 0. Which of the following best identifies the fundamental flaw in this reasoning?

A.The student failed to verify that terms are positive.
B.The student ignored the requirement that the sequence of absolute values must be monotonically decreasing. βœ…
C.The limit condition is sufficient for convergence, but the student did not check absolute convergence.
D.The series must start at n=1 rather than n=0 for the test to apply.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The Alternating Series Test requires two distinct conditions: the limit of the terms must be zero AND the magnitude of the terms must be monotonically non-increasing. A common misconception is assuming the limit condition alone guarantees convergence. However, if the terms do not decrease monotonically (e.g., they oscillate in magnitude while approaching zero), the partial sums may fail to settle toward a specific limit, causing divergence despite the individual terms vanishing.

Q2. Consider an alternating series where the magnitude of terms ana_n decreases to zero, but not monotonically (e.g., a2k<a2k+1a_{2k} < a_{2k+1}). What can be definitively concluded about its convergence?

A.It definitely diverges by the Alternating Series Test.
B.It converges conditionally but not absolutely.
C.The Alternating Series Test is inconclusive, and another method is required to determine convergence. βœ…
D.It converges absolutely because the terms approach zero.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: The Alternating Series Test provides sufficient conditions for convergence, not necessary ones. If the monotonicity condition fails, the test cannot be applied to prove convergence, nor does it automatically imply divergence. The series might still converge through other mechanisms or diverge due to the lack of cancellation. Therefore, one must employ alternative tests like the Ratio Test, Root Test, or direct analysis of partial sums to determine the behavior of such series.

Q3. When approximating the sum SS of a convergent alternating series using the partial sum sns_n, which statement correctly describes the relationship between the true sum and the approximation?

A.The error ∣Sβˆ’sn∣|S - s_n| is always exactly equal to the next term an+1a_{n+1}.
B.The true sum SS always lies strictly between sns_n and sn+1s_{n+1}, and the error is bounded by an+1a_{n+1}. βœ…
C.The error bound depends on the integral of the function generating the terms.
D.The approximation sns_n is always an overestimate if nn is odd.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: For a series satisfying the Alternating Series Test, the partial sums oscillate around the true sum SS. Specifically, SS is always trapped between any two consecutive partial sums sns_n and sn+1s_{n+1}. Consequently, the absolute error incurred by stopping at sns_n is guaranteed to be less than or equal to the magnitude of the first omitted term, an+1a_{n+1}. This property makes alternating series particularly useful for numerical approximation with controlled precision.

Q4. Given the series βˆ‘n=1∞(βˆ’1)nnn2+1\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 1}, a student applies the Alternating Series Test and concludes convergence. However, they also claim the series converges absolutely. Identify the error in this analysis.

A.The series actually diverges because the terms do not approach zero.
B.The series converges conditionally, not absolutely, because βˆ‘nn2+1\sum \frac{n}{n^2+1} behaves like the harmonic series. βœ…
C.The student should have used the Ratio Test instead of the Alternating Series Test.
D.There is no error; the series converges absolutely.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: While the alternating series converges because nn2+1\frac{n}{n^2+1} decreases to zero, absolute convergence requires βˆ‘nn2+1\sum \frac{n}{n^2+1} to converge. By Limit Comparison with βˆ‘1n\sum \frac{1}{n}, the series of absolute values diverges since the limit of the ratio is 1. Thus, the original series is only conditionally convergent. Confusing conditional and absolute convergence is a critical error because conditional convergence lacks properties like rearrangement invariance that absolute convergence possesses.

Q5. If the graph of the sequence of partial sums {sn}\{s_n\} for an alternating series shows oscillations that decrease in amplitude but never cross the horizontal axis after n=5n=5, what can be inferred about the sum SS?

A.The sum SS must be zero.
B.The sum SS has the same sign as s5s_5 and lies between 0 and s5s_5. βœ…
C.The series diverges because the partial sums stop crossing the axis.
D.The sum SS is negative regardless of the sign of s5s_5.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: In a convergent alternating series, partial sums alternate above and below the limit SS. If the oscillations cease crossing the axis after a certain point, it implies the limit SS itself acts as the boundary preventing further crossings. Since the amplitude decreases, all subsequent partial sums remain on one side of the axis, meaning SS must share the sign of those partial sums and lie between them and the axis (or between consecutive partial sums on that side).

Q6. Compare the efficiency of approximating ln⁑(2)\ln(2) using the standard alternating harmonic series versus the series derived from ln⁑(1+x1βˆ’x)\ln(\frac{1+x}{1-x}) with x=1/3x=1/3. Why is the latter preferred computationally?

A.The standard series has fewer terms.
B.The transformed series involves only even powers of x.
C.The transformed series converges geometrically fast due to the factor (1/3)2k+1(1/3)^{2k+1}, whereas the standard series converges harmonically slow. βœ…
D.The standard series is conditionally convergent while the transformed series is absolutely convergent.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: The standard alternating harmonic series converges very slowly; achieving kk decimal places requires roughly 10k10^k terms. In contrast, substituting x=1/3x=1/3 into 2(x+x3/3+… )2(x + x^3/3 + \dots) yields terms involving (1/9)k(1/9)^k, resulting in geometric convergence. Each additional term adds approximately two digits of accuracy. This dramatic difference in convergence rate makes the transformed series vastly superior for practical computation, illustrating how algebraic manipulation can optimize numerical methods derived from Taylor expansions.

Q7. A series βˆ‘(βˆ’1)nbn\sum (-1)^n b_n satisfies lim⁑bn=0\lim b_n = 0 and bn>0b_n > 0, but bn+1>bnb_{n+1} > b_n for infinitely many nn. Which of the following is a valid counterexample showing such a series can still converge?

A.bn=1n+(βˆ’1)nn2b_n = \frac{1}{n} + \frac{(-1)^n}{n^2} βœ…
B.bn=1nb_n = \frac{1}{n}
C.bn=1n+1n2b_n = \frac{1}{n} + \frac{1}{n^2}
D.No such series exists; monotonicity is necessary for convergence.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: This challenges the misconception that monotonicity is necessary. Consider bn=1n+(βˆ’1)nn2b_n = \frac{1}{n} + \frac{(-1)^n}{n^2}. While bnβ†’0b_n \to 0, it is not monotone due to the perturbation. However, splitting the series gives βˆ‘(βˆ’1)n/n\sum (-1)^n/n (convergent) plus βˆ‘1/n2\sum 1/n^2 (absolutely convergent). Their sum converges despite non-monotonicity. This demonstrates that the Alternating Series Test is sufficient but not necessary, and subtle structural properties can preserve convergence even when simple monotonicity fails.

Q8. In modeling a damped oscillating system, the displacement is given by βˆ‘(βˆ’1)ncneβˆ’knsin⁑(nt)\sum (-1)^n c_n e^{-kn} \sin(nt). If cnc_n grows polynomially, why does the series still converge for any fixed tt?

A.Polynomial growth dominates exponential decay.
B.Exponential decay eβˆ’kne^{-kn} dominates polynomial growth of cnc_n, ensuring terms decrease rapidly. βœ…
C.The sine function bounds the terms sufficiently.
D.The alternating signs cause perfect cancellation regardless of magnitude.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: In physical models combining oscillation and damping, convergence often hinges on competing growth rates. Even if coefficients cnc_n grow polynomially (e.g., npn^p), the exponential factor eβˆ’kne^{-kn} decays much faster. The product npeβˆ’knn^p e^{-kn} approaches zero monotonically for sufficiently large nn, satisfying the Alternating Series Test criteria eventually. This illustrates how asymptotic dominance determines convergence in applied contexts, overriding local irregularities or initial growth patterns in the coefficient sequence.

Q9. A student computes s10s_{10} for an alternating series and finds the error bound a11=0.004a_{11} = 0.004. They report the sum as 0.567Β±0.0040.567 \pm 0.004. Why might this interval estimate be overly conservative?

A.The actual error is typically much smaller than the first omitted term, especially if terms decrease rapidly. βœ…
B.The student should have used a10a_{10} instead of a11a_{11}.
C.Error bounds only apply to absolutely convergent series.
D.The sign of the error was not specified.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The Alternating Series Estimation Theorem provides a worst-case upper bound ∣Rnβˆ£β‰€an+1|R_n| \leq a_{n+1}. In practice, if terms decrease rapidly (e.g., factorially or geometrically), the remainder Rnβ‰ˆan+1βˆ’an+2+…R_n \approx a_{n+1} - a_{n+2} + \dots is significantly smaller than an+1a_{n+1} alone because subsequent terms partially cancel it. Reporting the full bound as uncertainty ignores this internal cancellation within the tail, potentially overstating the imprecision by an order of magnitude or more.

Q10. Which modification to the alternating harmonic series βˆ‘(βˆ’1)n+1/n\sum (-1)^{n+1}/n would result in a series that still converges but violates the monotonicity condition of the Alternating Series Test?

A.Multiplying each term by (1+1/n)(1 + 1/n)
B.Adding (βˆ’1)n/n2(-1)^n/n^2 to each term
C.Replacing nn with n+sin⁑(n)n + \sin(n) in the denominator βœ…
D.Doubling every even-indexed term
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Replacing nn with n+sin⁑(n)n + \sin(n) creates denominators that fluctuate, breaking strict monotonicity since sin⁑(n)\sin(n) is not monotone. However, the modified term behaves asymptotically like 1/n1/n with bounded perturbations. Using Dirichlet's Test or decomposing into βˆ‘(βˆ’1)n+1/n\sum (-1)^{n+1}/n plus a correction term reveals convergence persists. Options A and D alter asymptotic behavior or create divergence, while B maintains monotonicity for large n. This tests deep understanding beyond mechanical test application.

Q11. When applying the Alternating Series Test to βˆ‘(βˆ’1)nf(n)\sum (-1)^n f(n), a student verifies lim⁑f(n)=0\lim f(n)=0 but struggles to prove f(n)f(n) is decreasing algebraically. Which calculus-based strategy is most appropriate?

A.Check if f&#039;(x) < 0 for all xβ‰₯Nx \geq N where f(x)f(x) extends f(n)f(n). βœ…
B.Verify f(n+1)/f(n)<1f(n+1)/f(n) < 1 for all n.
C.Use L'HΓ΄pital's Rule on f(n)f(n).
D.Differentiate f(n)f(n) directly with respect to n.
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: When discrete monotonicity is hard to establish, extending f(n)f(n) to a continuous differentiable function f(x)f(x) allows using derivatives. If f&#039;(x) < 0 for xβ‰₯Nx \geq N, then f(x)f(x) is decreasing on that interval, implying f(n)f(n) is decreasing for nβ‰₯Nn \geq N. This bridges discrete sequences and continuous calculus, providing a powerful tool when algebraic comparison of successive terms becomes cumbersome. Note that differentiation with respect to discrete n is undefined, making option D invalid.

Q12. Consider two alternating series A=βˆ‘(βˆ’1)nanA = \sum (-1)^n a_n and B=βˆ‘(βˆ’1)nbnB = \sum (-1)^n b_n where both satisfy AST conditions. If an<bna_n < b_n for all n, which statement about their sums SAS_A and SBS_B is necessarily true?

A.SA<SBS_A < S_B
B.∣SA∣<∣SB∣|S_A| < |S_B|
C.No definitive ordering exists without knowing specific term structures. βœ…
D.SAS_A and SBS_B have opposite signs.
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: Unlike series with positive terms, alternating series sums depend critically on phase and cancellation patterns, not just term magnitudes. Even if an<bna_n < b_n, the interplay of signs means SAS_A could exceed, equal, or fall below SBS_B. For example, shifting indices or altering initial terms changes sums disproportionately. This highlights that magnitude comparisons of terms do not translate directly to sum comparisons for alternating series, distinguishing them fundamentally from positive-term series theory.

Q13. A computational algorithm approximates Ο€\pi using Leibniz's formula 4βˆ‘(βˆ’1)n/(2n+1)4\sum (-1)^n/(2n+1). After 1 million terms, accuracy is poor. Which transformation would most effectively accelerate convergence while preserving the alternating structure?

A.Euler's transform for alternating series βœ…
B.Simply computing more terms
C.Using the geometric series for 1/(1+x)1/(1+x)
D.Applying the Integral Test
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Leibniz's series converges extremely slowly (error ~ 1/n). Euler's transform specifically accelerates alternating series by reweighting partial sums based on finite differences, often converting logarithmic/harmonic convergence to geometric convergence without changing the underlying sum. While other transformations exist, Euler's is tailored for this structure. Merely adding terms is inefficient, geometric series don't directly apply, and Integral Test only assesses convergence. This represents advanced numerical analysis knowledge connecting series theory to computational efficiency.

Q14. If βˆ‘(βˆ’1)nan\sum (-1)^n a_n converges conditionally, what happens to the sum when terms are rearranged so that two positive terms alternate with one negative term?

A.The sum remains unchanged due to commutativity.
B.The new series diverges to infinity.
C.The sum changes to a different finite value. βœ…
D.The new series oscillates without converging.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Riemann's Rearrangement Theorem states that conditionally convergent series can be rearranged to converge to any real number or diverge. The specific pattern of 2 positives to 1 negative alters the balance of cancellation, shifting the limit to a new value (specifically 32ln⁑2\frac{3}{2}\ln 2 for alternating harmonic). This contrasts sharply with absolutely convergent series where rearrangements preserve sums. Understanding this fragility is crucial for recognizing why absolute convergence is structurally significant beyond mere convergence.

Q15. A student analyzes βˆ‘(βˆ’1)nln⁑nn\sum (-1)^n \frac{\ln n}{n} and claims AST fails because ln⁑nn\frac{\ln n}{n} increases initially. How should this be corrected?

A.AST requires global monotonicity from n=1, so the series diverges.
B.AST only requires eventual monotonicity; since derivative becomes negative for n>en > e, convergence holds. βœ…
C.The series diverges because ln n grows without bound.
D.Logarithmic terms invalidate AST entirely.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: The Alternating Series Test requires monotonic decrease only eventually (for nβ‰₯Nn \geq N), not from the start. The function f(x)=ln⁑x/xf(x) = \ln x / x has derivative (1βˆ’ln⁑x)/x2(1-\ln x)/x^2, which is negative for x>ex > e. Thus, terms decrease for nβ‰₯3n \geq 3. Initial non-monotonicity affects only finitely many terms and doesn't impact convergence. This corrects the common misconception that AST demands strict monotonicity from the first term, emphasizing asymptotic behavior over initial transients.

Q16. In the context of power series evaluated at endpoints, why does the alternating series test frequently determine convergence at x=Rx = R when the ratio test fails?

A.The ratio test always gives ρ=1 at endpoints, while AST exploits sign alternation and term decay specific to that point. βœ…
B.Power series always converge at endpoints if alternating.
C.The ratio test cannot handle negative terms.
D.AST is stronger than the ratio test everywhere.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: At radius boundaries, the ratio/root tests yield ρ=1 (inconclusive). However, substituting x=Rx=R often produces an alternating numerical series whose terms decay appropriately. AST leverages this specific structureβ€”alternation plus decayβ€”to establish convergence where magnitude-based tests fail. This synergy explains why endpoint analysis typically shifts from ratio tests to AST or p-series tests. It underscores that convergence at boundaries depends on delicate cancellations invisible to pure magnitude assessments.

Q17. Suppose βˆ‘(βˆ’1)nan\sum (-1)^n a_n satisfies AST. If we define bn=anb_n = a_n for odd n and bn=an/2b_n = a_n/2 for even n, does βˆ‘(βˆ’1)nbn\sum (-1)^n b_n necessarily converge?

A.Yes, because bn→0b_n \to 0 and alternates.
B.No, because monotonicity is destroyed by halving even terms. βœ…
C.Yes, because it is dominated by the original convergent series.
D.Convergence depends on whether ana_n decreases fast enough.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Halving even terms creates a sawtooth pattern where b2k=a2k/2<a2k+1=b2k+1b_{2k} = a_{2k}/2 < a_{2k+1} = b_{2k+1} potentially, violating monotonicity. Unlike the earlier counterexample where perturbation preserved convergence via decomposition, here the systematic reduction of every other term disrupts the cancellation mechanism essential to AST. The new series may diverge because the imbalance prevents partial sums from stabilizing. This demonstrates that arbitrary modifications to term magnitudes can destroy convergence even when individual terms still vanish, highlighting the delicacy of conditional convergence.

Q18. When estimating ∫01sin⁑xxdx\int_0^1 \frac{\sin x}{x} dx via series, why is the alternating series error bound preferable to Taylor's remainder formula?

A.Taylor's formula requires bounding high-order derivatives which is complex for sin(x)/x.
B.The integrated series is alternating with rapidly decreasing terms, making an+1a_{n+1} a simpler tight bound. βœ…
C.Both methods give identical bounds always.
D.Taylor's formula doesn't apply to integrals.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Expanding sin⁑x/x=1βˆ’x2/3!+x4/5!βˆ’β€¦\sin x/x = 1 - x^2/3! + x^4/5! - \dots and integrating yields an alternating series with factorial-denominated terms. Applying AST error bound ∣Rnβˆ£β‰€an+1|R_n| \leq a_{n+1} avoids computing messy derivatives of sin⁑x/x\sin x/x needed for Lagrange remainder. The factorial decay ensures the first omitted term is an excellent error proxy. This showcases practical advantage: when integration produces alternating series naturally, AST provides computationally simpler and often tighter error control than general Taylor estimates.

Q19. A series βˆ‘(βˆ’1)nan\sum (-1)^n a_n has terms satisfying anβ†’0a_n \to 0 and an+1≀ana_{n+1} \leq a_n for all n except n=100n=100. Can AST be applied?

A.No, exceptions invalidate the test completely.
B.Yes, because finite exceptions don't affect convergence or applicability of AST. βœ…
C.Only if a100<a99a_{100} < a_{99} also holds.
D.AST requires checking all terms individually.
πŸ’‘ Difficulty: medium | βœ… Correct: B

Q20. Consider βˆ‘(βˆ’1)n(n+1βˆ’n)\sum (-1)^n (\sqrt{n+1} - \sqrt{n}). Without rationalizing, a student claims divergence because n+1>n\sqrt{n+1} > \sqrt{n}. What is the correct analysis?

A.The student is correct; increasing terms imply divergence.
B.After rationalization, terms become 1n+1+n\frac{1}{\sqrt{n+1}+\sqrt{n}} which decreases to 0, so AST confirms convergence. βœ…
C.The series diverges by nth term test.
D.Terms increase but approach 0, so AST still applies directly.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: The raw form n+1βˆ’n\sqrt{n+1}-\sqrt{n} appears problematic for monotonicity assessment. Rationalizing reveals equivalent form 1/(n+1+n)1/(\sqrt{n+1}+\sqrt{n}), clearly decreasing to 0. The student misidentified term behavior by focusing on components rather than the difference. This exemplifies why algebraic simplification precedes test application: apparent non-monotonicity may be illusory. The series actually converges by AST after proper manipulation, correcting the superficial analysis that mistook component growth for term growth.

Q21. In a physics model, force F=βˆ‘(βˆ’1)ncnrnF = \sum (-1)^n c_n r^n converges for r<Rr < R. At r=Rr = R, terms alternate and cnRnβ†’0c_n R^n \to 0 monotonically. What physical implication does convergence at the boundary have?

A.The model is valid exactly at the critical radius R. βœ…
B.The model breaks down at R regardless of mathematical convergence.
C.Convergence at R implies absolute convergence inside.
D.Physical quantities cannot be represented by alternating series.
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: Mathematical convergence at boundary r=Rr=R via AST validates the series representation precisely at that critical point, extending the domain of validity beyond open interval (βˆ’R,R)(-R,R). Physically, this means the modeled quantity remains well-defined and finite at the threshold where naive radius tests are inconclusive. This bridges abstract analysis and applied modeling: proving boundary convergence confirms physical predictability at critical parameters, whereas divergence would signal model breakdown or phase transition requiring alternative formulations.

Q22. Why can't the Alternating Series Test be used to prove divergence of βˆ‘(βˆ’1)nnn+1\sum (-1)^n \frac{n}{n+1}?

A.Because the terms don't alternate properly.
B.Because AST is a convergence test only; failure of conditions doesn't imply divergence. βœ…
C.Because the series actually converges.
D.Because divergence requires the Ratio Test.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: AST provides sufficient conditions for convergence, not necessary conditions for divergence. When lim⁑anβ‰ 0\lim a_n \neq 0, divergence follows from nth Term Test, not AST failure per se. But if monotonicity fails while limit is zero, AST simply doesn't applyβ€”it doesn't declare divergence. Students often mistakenly treat failed AST as divergence proof. Correct approach: use nth Term Test for non-zero limits, other tests for zero-limit non-monotone cases. Understanding test scope prevents logical errors in series classification.

Q23. Given S=βˆ‘n=1∞(βˆ’1)n+1/n2S = \sum_{n=1}^\infty (-1)^{n+1}/n^2, and knowing βˆ‘1/n2=Ο€2/6\sum 1/n^2 = \pi^2/6, find SS without direct summation.

A.Ο€2/12\pi^2/12 βœ…
B.Ο€2/6\pi^2/6
C.βˆ’Ο€2/12-\pi^2/12
D.Cannot be determined from given information.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Decompose βˆ‘(βˆ’1)n+1/n2=βˆ‘1/n2βˆ’2βˆ‘1/(2k)2=Ο€2/6βˆ’2(1/4)βˆ‘1/k2=Ο€2/6βˆ’Ο€2/12=Ο€2/12\sum (-1)^{n+1}/n^2 = \sum 1/n^2 - 2\sum 1/(2k)^2 = \pi^2/6 - 2(1/4)\sum 1/k^2 = \pi^2/6 - \pi^2/12 = \pi^2/12. This uses relationship between alternating and absolute series: Salt=Sabsβˆ’2SevenS_{alt} = S_{abs} - 2S_{even}. Recognizing this structural link avoids re-summing and leverages known results. It demonstrates how alternating series connect to their absolute counterparts, enabling efficient evaluation through algebraic decomposition rather than brute-force computation or advanced special functions.

Q24. A student uses AST on βˆ‘(βˆ’1)nsin⁑(1/n)\sum (-1)^n \sin(1/n) and argues monotonicity holds because sin x is increasing. Identify the error.

A.Sin x is increasing, but sin(1/n) decreases as n increases since 1/n decreases. βœ…
B.The argument is correct; no error exists.
C.AST doesn't apply to trigonometric terms.
D.Monotonicity requires derivative test, not function properties.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: The student confused monotonicity of sin(x) with monotonicity of composite sin(1/n). Since 1/n is decreasing and sin is increasing on (0,1], the composition sin(1/n) is indeed decreasing. However, the student's reasoning cited sin x increasing as justification, which is backwards logic even though conclusion happens to be correct. Proper reasoning: outer function increasing + inner function decreasing β‡’ composite decreasing. This highlights need for precise chain-rule thinking in sequence monotonicity, not just recalling elementary function behaviors in isolation.

Q25. For the series βˆ‘(βˆ’1)nan\sum (-1)^n a_n where an=∫nn+1dxxa_n = \int_n^{n+1} \frac{dx}{x}, explain why AST applies without explicit formula for ana_n.

A.Integrals always produce decreasing sequences.
B.Since 1/x1/x is decreasing, ∫nn+1dxx>∫n+1n+2dxx\int_n^{n+1} \frac{dx}{x} > \int_{n+1}^{n+2} \frac{dx}{x}, ensuring monotonicity; limit is 0 by squeeze theorem. βœ…
C.AST applies to all integral-defined sequences.
D.The integral equals ln(1+1/n) which obviously decreases.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Without evaluating the integral, note f(x)=1/xf(x)=1/x decreasing implies ∫nn+1f>∫n+1n+2f\int_n^{n+1} f > \int_{n+1}^{n+2} f, establishing monotonicity directly from integrand properties. Also 0<an<1/nβ†’00 < a_n < 1/n \to 0. Thus AST conditions satisfied purely analytically. This demonstrates powerful technique: leveraging integral properties to verify series conditions when closed forms are unavailable or messy. It connects integral calculus and series convergence structurally, avoiding explicit antiderivative computation while rigorously justifying test applicability.

Q26. If βˆ‘(βˆ’1)nan\sum (-1)^n a_n converges and an>0a_n > 0, which statement about βˆ‘(βˆ’1)nan2\sum (-1)^n a_n^2 is always true?

A.It converges absolutely.
B.It converges conditionally.
C.It diverges.
D.No universal conclusion; depends on decay rate of ana_n. βœ…
πŸ’‘ Difficulty: hard | βœ… Correct: D

Q27. In approximating eβˆ’1e^{-1} via βˆ‘(βˆ’1)n/n!\sum (-1)^n/n!, why does the error bound 1/(n+1)!1/(n+1)! become accurate extremely quickly compared to geometric series error bounds?

A.Factorial decay dominates exponential/geometric decay, making omitted terms negligible sooner. βœ…
B.Geometric series have larger constants.
C.Error bounds for geometric series are inherently loose.
D.eβˆ’1e^{-1} is a special number with unique properties.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Factorial denominators n!n! grow super-exponentially, far outpacing geometric rnr^n. Thus 1/(n+1)!1/(n+1)! plummets faster than any geometric bound CrnCr^n. For n=10, 1/11!β‰ˆ2.5Γ—10βˆ’81/11! \approx 2.5\times10^{-8} vs geometric (0.5)10β‰ˆ10βˆ’3(0.5)^{10} \approx 10^{-3}. This explains why Taylor series for entire functions like exe^x achieve machine precision in few terms, while geometric approximations require many more. Understanding decay hierarchy (factorial > exponential > polynomial) is key to selecting efficient approximation methods in numerical analysis.

Q28. A series satisfies anβ†’0a_n \to 0 and alternates, but an+1≀ana_{n+1} \leq a_n fails for a sparse subsequence nk=2kn_k = 2^k. Does this prevent AST application?

A.Yes, any failure invalidates AST.
B.No, if monotonicity holds for all other n and violations don't accumulate destructively, convergence may still hold via refined analysis. βœ…
C.AST requires checking only prime indices.
D.The series automatically diverges.
πŸ’‘ Difficulty: hard | βœ… Correct: B

Q29. When graphing partial sums sns_n of a convergent alternating series, the envelope curves bounding oscillations correspond to what mathematical objects?

A.The functions f(x)f(x) and βˆ’f(x)-f(x) where f(n)=anf(n)=a_n. βœ…
B.Horizontal lines at Β±S.
C.Exponential decay curves unrelated to terms.
D.Parabolas fitted to extrema.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Partial sums oscillate within envelopes defined by cumulative effect of term magnitudes. For well-behaved alternating series, upper/lower bounds of oscillation trace curves related to term sequence ana_n and its integral/sum analogs. Specifically, sns_n stays between curves reflecting running totals of positive/negative contributions, often approximated by SΒ±anS \pm a_n locally. Visualizing these envelopes helps diagnose convergence rate and error behavior intuitively, connecting discrete partial sum geometry to continuous term decay profiles without heavy computation.

Q30. Why is conditional convergence of alternating series problematic in computer floating-point arithmetic?

A.Computers cannot represent irrational sums.
B.Finite precision destroys exact cancellation, accumulating rounding errors that swamp true sum for slowly convergent series. βœ…
C.Alternating series require infinite memory.
D.Computers only handle absolutely convergent series.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Conditional convergence relies on precise cancellation between positive and negative terms. Floating-point rounding introduces tiny errors that don't cancel perfectly, accumulating systematically. For slowly convergent alternating series (e.g., harmonic), millions of terms amplify these errors until they dominate the true sum. Absolutely convergent series are robust because term magnitudes decay fast enough that rounding errors remain bounded. This practical limitation explains why numerically stable algorithms avoid naive summation of conditionally convergent series, preferring accelerated or rearranged forms.

Q31. If βˆ‘(βˆ’1)nan\sum (-1)^n a_n converges by AST and βˆ‘an\sum a_n diverges, what can be said about βˆ‘(βˆ’1)nan/ln⁑n\sum (-1)^n a_n / \ln n?

A.It diverges because dividing by ln n slows decay.
B.It converges absolutely.
C.It converges conditionally by AST since an/ln⁑na_n/\ln n inherits monotonicity and limit 0. βœ…
D.Insufficient information.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Since ana_n decreases to 0 and ln⁑n\ln n increases, an/ln⁑na_n/\ln n also decreases to 0 (product of decreasing positive and reciprocal of increasing positive). Thus AST applies directly, guaranteeing convergence. Absolute convergence fails since an/ln⁑n>an/na_n/\ln n > a_n/n potentially and βˆ‘an\sum a_n already diverges. So conditional convergence persists. This shows robustness of AST under certain transformations: dividing by slowly growing functions preserves the structural conditions needed for alternating convergence even when absolute convergence is impossible.

Q32. A student asserts βˆ‘(βˆ’1)nn!nn\sum (-1)^n \frac{n!}{n^n} diverges because factorial grows fast. Correct this using AST.

A.Factorial growth in numerator causes divergence.
B.Ratio test shows ρ=1/e<1\rho = 1/e < 1, so terms decrease rapidly to 0; AST confirms convergence. βœ…
C.AST doesn't apply to factorials.
D.The series is absolutely convergent, so AST is irrelevant.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Student confused numerator growth with term behavior. Actually n!/nnβ†’0n!/n^n \to 0 super-exponentially (Stirling/Ratio Test: ρ=1/e\rho=1/e). Terms decrease monotonically for large n. Thus AST readily confirms convergenceβ€”in fact absolute convergence holds. The error stems from misjudging asymptotic dominance: denominator nnn^n overwhelms n!n!. This corrects intuitive but wrong heuristic about factorials, emphasizing rigorous limit/ratio analysis over superficial growth impressions when assessing series term behavior.

Q33. In the series βˆ‘(βˆ’1)narctan⁑nn\sum (-1)^n \frac{\arctan n}{n}, why does AST apply despite arctan n approaching Ο€/2 β‰  0?

A.It doesn't apply; series diverges by nth term test.
B.Arctan n / n β†’ 0 because denominator dominates; monotonicity holds since derivative negative for large n. βœ…
C.Arctan is bounded so series converges trivially.
D.AST applies to any bounded numerator.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Although arctan n β†’ Ο€/2, division by n drives term to 0. Monotonicity: derivative of arctan x / x is x/(1+x2)βˆ’arctan⁑xx2<0\frac{x/(1+x^2) - \arctan x}{x^2} < 0 for x>0 since arctan x > x/(1+xΒ²). Thus terms decrease to 0, satisfying AST. Key insight: bounded numerator divided by unbounded denominator yields vanishing terms; derivative analysis confirms monotonicity despite non-obvious numerator behavior. This combines limit arithmetic with calculus verification for non-trivial AST application.

Q34. Suppose you know βˆ‘(βˆ’1)nan=S\sum (-1)^n a_n = S and ana_n satisfies AST. If you compute sns_n and find sn>Ss_n > S, what must be true about n?

A.n is even.
B.n is odd. βœ…
C.n is greater than 100.
D.Cannot determine parity from single inequality.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: For standard alternating series starting with positive term (a1>0a_1 > 0), odd partial sums s2kβˆ’1s_{2k-1} overestimate S while even partial sums s2ks_{2k} underestimate S. Thus sn>Ss_n > S implies n is odd. This geometric property arises from oscillation pattern: adding positive term overshoots, subtracting undershoots. Recognizing this parity-sum relationship enables error sign prediction and validation of computational results without recomputation, linking visual oscillation behavior to index parity systematically.

Q35. Why can't we conclude βˆ‘(βˆ’1)nanbn\sum (-1)^n a_n b_n converges just because βˆ‘(βˆ’1)nan\sum (-1)^n a_n converges by AST and bnβ†’0b_n \to 0?

A.Product of convergent series always diverges.
B.bnb_n might oscillate or grow temporarily, destroying monotonicity of product even if both factors behave well individually. βœ…
C.AST requires bn=1b_n = 1 always.
D.Convergence of product requires absolute convergence of both.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Even if an↓0a_n \downarrow 0 and bnβ†’0b_n \to 0, the product anbna_n b_n may not be monotone. Example: an=1/na_n = 1/n, bn=(2+(βˆ’1)n)/nb_n = (2 + (-1)^n)/n. Product oscillates in magnitude, violating AST monotonicity. Convergence isn't guaranteed without additional constraints (e.g., bnb_n monotone or bounded variation). This illustrates that convergence properties aren't multiplicative; structural conditions like monotonicity are fragile under multiplication, requiring careful joint analysis rather than separate verification.

Q36. An alternating series model for population dynamics predicts extinction if sum < 0. Given Sβ‰ˆs50=βˆ’0.001S \approx s_{50} = -0.001 with error bound a51=0.002a_{51} = 0.002, can extinction be confirmed?

A.Yes, since approximation is negative.
B.No, because error bound exceeds magnitude of approximation; true sum could be positive. βœ…
C.Extinction depends on initial conditions, not series sum.
D.More terms are needed but extinction is likely.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: With s50=βˆ’0.001s_{50} = -0.001 and error ≀ 0.002, true sum lies in [-0.003, 0.001]. Since interval includes positive values, extinction (S<0) cannot be conclusively determined from this approximation. The uncertainty spans the decision threshold. This demonstrates critical importance of error bounds in applied modeling: numerical proximity to zero isn't sufficient when tolerance exceeds signal magnitude. Practical decisions require either tighter bounds (more terms) or analytical proof of sign, highlighting intersection of numerical analysis and real-world decision-making under uncertainty.

Q37. Which series transformation converts βˆ‘(βˆ’1)n/n\sum (-1)^n / n into a faster-converging alternating series without changing its sum?

A.Abel summation
B.Euler transform βœ…
C.CesΓ ro summation
D.Borel summation
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Euler transform specifically targets alternating series, recombining terms via binomial weights to accelerate convergence while preserving sum. For alternating harmonic, it dramatically improves rate. Abel/CesΓ ro/Borel address divergent or summability issues, not acceleration of convergent alternating series. This specialized technique exploits alternating structure uniquely, unlike general summability methods. Knowledge of such transforms represents advanced series manipulation skill beyond standard curriculum, connecting classical analysis to modern computational mathematics for optimizing slowly convergent alternating series encountered in research and high-precision applications.

πŸ”— Related Topics (MCQs)