Calculus 2 Quiz: Radius And Interval Of Convergence
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Radius And Interval Of ConvergenceQuestion 1 of 20

Find the interval of convergence of the series ∑n=1∞(2x−5)nn2\sum_{n=1}^{\infty} \frac{(2x-5)^n}{n^2}

(2,3)(2, 3)
[2,3][2, 3]
(−1,1)(-1, 1)
(4,6)(4, 6)
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Calculus 2 Quiz

Calculus 2 Quiz: Radius And Interval Of Convergence

Practice Radius And Interval Of Convergence in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Radius And Interval Of Convergence, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

Find the interval of convergence of the series ∑n=1∞(2x−5)nn2\sum_{n=1}^{\infty} \frac{(2x-5)^n}{n^2}

  1. (2,3)(2, 3)
  2. [2,3][2, 3] (correct answer)
  3. (−1,1)(-1, 1)
  4. (4,6)(4, 6)
Explanation: Let u=2x−5u = 2x-5. The series becomes ∑n=1∞unn2\sum_{n=1}^{\infty} \frac{u^n}{n^2}. The radius of convergence for this series in uu is Ru=1R_u=1, converging for ∣u∣<1|u|<1. Substituting back, we have ∣2x−5∣<1|2x-5|<1, which is −1<2x−5<1-1 < 2x-5 < 1. Adding 5 gives 4<2x<64 < 2x < 6, and dividing by 2 gives 2<x<32 < x < 3. Now we check the endpoints. At x=2x=2, u=−1u=-1 and the series is ∑(−1)nn2\sum \frac{(-1)^n}{n^2}, which converges absolutely. At x=3x=3, u=1u=1 and the series is ∑1n2\sum \frac{1}{n^2}, a convergent p-series. Both endpoints are included, so the interval is [2,3][2, 3].

Question 2

What is the radius of convergence for the power series ∑n=1∞(n+12n+3)n(x+1)n\sum_{n=1}^{\infty} \left(\frac{n+1}{2n+3}\right)^n (x+1)^n?

  1. R=1/2R = 1/2
  2. R=1R = 1
  3. R=2R = 2 (correct answer)
  4. R=∞R = \infty
Explanation: Using the Root Test, we evaluate the limit L=lim⁡n→∞∣(n+12n+3)n(x+1)n∣n=lim⁡n→∞∣n+12n+3(x+1)∣L = \lim_{n \to \infty} \sqrt[n]{\left| \left(\frac{n+1}{2n+3}\right)^n (x+1)^n \right|} = \lim_{n \to \infty} \left| \frac{n+1}{2n+3} (x+1) \right|. This simplifies to L=∣12(x+1)∣L = \left| \frac{1}{2}(x+1) \right|. For convergence, we require L<1L < 1, so ∣x+12∣<1\left| \frac{x+1}{2} \right| < 1, which implies ∣x+1∣<2|x+1| < 2. By the definition ∣x−c∣<R|x-c| < R, the radius of convergence is R=2R=2.

Question 3

Find the radius of convergence for the power series ∑n=1∞(1+1n)nln⁡nxn\sum_{n=1}^\infty \left(1 + \frac{1}{n}\right)^{n \ln n} x^n

  1. R=0R = 0
  2. R=1R = 1 (correct answer)
  3. R=eR = e
  4. R=∞R = \infty
Explanation: Using the Root Test, we evaluate L=lim⁡n→∞∣(1+1n)nln⁡nxn∣n=∣x∣lim⁡n→∞(1+1n)ln⁡nL = \lim_{n \to \infty} \sqrt[n]{\left| \left(1 + \frac{1}{n}\right)^{n \ln n} x^n \right|} = |x| \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\ln n}. To evaluate the limit, let y=(1+1/n)ln⁡ny = (1 + 1/n)^{\ln n}, so ln⁡y=(ln⁡n)ln⁡(1+1/n)\ln y = (\ln n) \ln(1+1/n). Using the approximation ln⁡(1+u)≈u\ln(1+u) \approx u for small uu, we get ln⁡y≈(ln⁡n)(1/n)=ln⁡nn\ln y \approx (\ln n)(1/n) = \frac{\ln n}{n}. As n→∞n \to \infty, lim⁡ln⁡nn=0\lim \frac{\ln n}{n} = 0. So lim⁡ln⁡y=0\lim \ln y = 0, which means lim⁡y=e0=1\lim y = e^0 = 1. Thus, L=∣x∣⋅1=∣x∣L = |x| \cdot 1 = |x|. For convergence, ∣x∣<1|x|<1, so the radius of convergence is R=1R=1.

Question 4

Consider the power series ∑n=1∞(x+2)nn⋅4n\sum_{n=1}^{\infty} \frac{(x+2)^n}{n \cdot 4^n}. At which of the following points does the series converge?

  1. x=2x = 2 only
  2. x=−6x = -6 only (correct answer)
  3. x=2x = 2 and x=−6x = -6
  4. Neither x=2x = 2 nor x=−6x = -6
Explanation: Using ratio test: lim⁡n→∞∣n(n+1)⋅4∣=14\lim_{n \to \infty} \left|\frac{n}{(n+1) \cdot 4}\right| = \frac{1}{4}, so R=4R = 4. The series converges for ∣x+2∣<4|x+2| < 4, i.e., −6<x<2-6 < x < 2. At x=2x = 2: ∑n=1∞4nn⋅4n=∑n=1∞1n\sum_{n=1}^{\infty} \frac{4^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{1}{n} (harmonic series), which diverges. At x=−6x = -6: ∑n=1∞(−4)nn⋅4n=∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-4)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} (alternating harmonic series), which converges. Therefore, the series converges at x=−6x = -6 only.

Question 5

The coefficients of a power series ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n satisfy the recurrence relation an+1=n2n+5ana_{n+1} = \frac{n}{2n+5} a_n for n≥0n \ge 0, with a0=1a_0=1. Find the radius of convergence of the series.

  1. R=1/2R = 1/2
  2. R=2R = 2 (correct answer)
  3. R=5/2R = 5/2
  4. R=1R = 1
Explanation: We can use the Ratio Test directly on the definition of the series. The limit of the ratio of consecutive terms is L=lim⁡n→∞∣an+1xn+1anxn∣=∣x∣lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| = |x| \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. Using the given recurrence relation, an+1an=n2n+5\frac{a_{n+1}}{a_n} = \frac{n}{2n+5}. The limit is lim⁡n→∞n2n+5=12\lim_{n \to \infty} \frac{n}{2n+5} = \frac{1}{2}. So, L=12∣x∣L = \frac{1}{2}|x|. The series converges when L<1L < 1, which means 12∣x∣<1\frac{1}{2}|x| < 1, or ∣x∣<2|x| < 2. The radius of convergence is R=2R=2.

Question 6

The power series ∑n=2∞xnn(ln⁡n)2\sum_{n=2}^{\infty} \frac{x^n}{n(\ln n)^2} has interval of convergence II. Which statement about II is correct?

  1. I=[−1,1]I = [-1, 1] and the series converges absolutely at both endpoints (correct answer)
  2. I=[−1,1]I = [-1, 1] and the series converges conditionally at both endpoints
  3. I=(−1,1]I = (-1, 1] and the series converges conditionally at x=1x = 1
  4. I=[−1,1)I = [-1, 1) and the series converges conditionally at x=−1x = -1
Explanation: Using ratio test: lim⁡n→∞∣n(ln⁡n)2(n+1)(ln⁡(n+1))2∣=1\lim_{n \to \infty} \left|\frac{n(\ln n)^2}{(n+1)(\ln(n+1))^2}\right| = 1, so R=1R = 1. At x=1x = 1: ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} converges by integral test since ∫2∞1x(ln⁡x)2dx=[−1ln⁡x]2∞=1ln⁡2\int_2^{\infty} \frac{1}{x(\ln x)^2} dx = \left[-\frac{1}{\ln x}\right]_2^{\infty} = \frac{1}{\ln 2}. At x=−1x = -1: ∑n=2∞(−1)nn(ln⁡n)2\sum_{n=2}^{\infty} \frac{(-1)^n}{n(\ln n)^2} converges absolutely since ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} converges. Therefore I=[−1,1]I = [-1, 1] with absolute convergence at both endpoints.

Question 7

For the power series ∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}, which statement correctly describes its interval of convergence?

  1. The series converges for ∣x∣<1|x| < 1 and diverges at x=±1x = \pm 1
  2. The series converges for all real numbers xx (correct answer)
  3. The series converges for ∣x∣≤1|x| \leq 1 and diverges for ∣x∣>1|x| > 1
  4. The series converges for x∈(−∞,0]∪[0,∞)x \in (-\infty, 0] \cup [0, \infty) with radius R=∞R = \infty
Explanation: Using the ratio test on the series ∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}: lim⁡n→∞∣x2(n+1)(2n)!x2n(2(n+1))!∣=lim⁡n→∞x2(2n+2)(2n+1)=0\lim_{n \to \infty} \left|\frac{x^{2(n+1)}(2n)!}{x^{2n}(2(n+1))!}\right| = \lim_{n \to \infty} \frac{x^2}{(2n+2)(2n+1)} = 0 for any finite xx. Since this limit is 0 (less than 1) for all xx, the radius of convergence is R=∞R = \infty. This series is actually cos⁡(x)\cos(x), which converges for all real xx. Choice A assumes finite radius, choice C gives the wrong interval, and choice D uses confusing notation that essentially says the same as choice B but in a misleading way.

Question 8

Let RR be the radius of convergence of the power series ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n. If it is known that the series of coefficients ∑n=0∞an\sum_{n=0}^{\infty} a_n converges, what is the strongest conclusion that can be made about RR?

  1. R=1R = 1
  2. R>1R > 1
  3. R≤1R \le 1
  4. R≥1R \ge 1 (correct answer)
Explanation: The series ∑n=0∞an\sum_{n=0}^{\infty} a_n is the power series ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n evaluated at x=1x=1. The fact that the power series converges for a specific value x=1x=1 means that its radius of convergence RR must be at least the distance from the center (c=0) to that point. So, R≥∣1−0∣=1R \ge |1-0| = 1. It is possible that R=1R=1 (e.g., for an=(−1)n/na_n = (-1)^n/n) or that R>1R>1 (e.g., for an=(1/2)na_n = (1/2)^n). The only certain conclusion is R≥1R \ge 1.

Question 9

The power series f(x)=∑n=1∞(x−4)nn2f(x) = \sum_{n=1}^{\infty} \frac{(x-4)^n}{n^2} has an interval of convergence of [3,5][3, 5]. What is the interval of convergence for the series representing its derivative, f′(x)f'(x)?

  1. (3,5)(3, 5)
  2. [3,5][3, 5]
  3. [3,5)[3, 5) (correct answer)
  4. (3,5](3, 5]
Explanation: Differentiation of a power series does not change its radius of convergence. The series for f(x)f(x) is centered at c=4c=4 and has R=1R=1, so the series for f′(x)f'(x) also has R=1R=1 and converges on (3,5)(3, 5). We must re-check the endpoints for the derivative series, f′(x)=∑n=1∞n(x−4)n−1n2=∑n=1∞(x−4)n−1nf'(x) = \sum_{n=1}^{\infty} \frac{n(x-4)^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{(x-4)^{n-1}}{n}. At x=5x=5, the series is ∑1n\sum \frac{1}{n}, the harmonic series, which diverges. At x=3x=3, the series is ∑(−1)n−1n\sum \frac{(-1)^{n-1}}{n}, the alternating harmonic series, which converges. Thus, the interval of convergence for f′(x)f'(x) is [3,5)[3, 5).

Question 10

The interval of convergence of the power series ∑n=1∞an(x−2)n\sum_{n=1}^{\infty} a_n (x-2)^n is (−1,5](-1, 5]. Which of the following series must converge? \ I. ∑n=1∞an\sum_{n=1}^{\infty} a_n \ II. ∑n=1∞an3n\sum_{n=1}^{\infty} a_n 3^n \ III. ∑n=1∞an(−3)n\sum_{n=1}^{\infty} a_n (-3)^n

  1. I only
  2. II only
  3. I and II only (correct answer)
  4. I, II, and III
Explanation: We test which value of xx corresponds to each series. For series I, ∑an\sum a_n, we need (x−2)n=1n(x-2)^n = 1^n, so x−2=1x-2=1 and x=3x=3. Since x=3x=3 is inside the interval (−1,5](-1, 5], series I must converge. For series II, ∑an3n\sum a_n 3^n, we need (x−2)n=3n(x-2)^n = 3^n, so x−2=3x-2=3 and x=5x=5. Since the interval includes the endpoint x=5x=5, series II must converge. For series III, ∑an(−3)n\sum a_n (-3)^n, we need (x−2)n=(−3)n(x-2)^n = (-3)^n, so x−2=−3x-2=-3 and x=−1x=-1. Since the interval (−1,5](-1, 5] excludes the endpoint x=−1x=-1, series III must diverge. Therefore, only series I and II must converge.

Question 11

The Maclaurin series for cos⁡(x)\cos(x) is ∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}, which has an infinite radius of convergence. What is the interval of convergence for the Taylor series of f(x)=cos⁡(x−1)f(x)=\cos(x-1) centered at c=1c=1?

  1. (0,2)(0, 2)
  2. [−1,1][-1, 1]
  3. (−∞,∞)(-\infty, \infty) (correct answer)
  4. Only for x=1x=1.
Explanation: The Taylor series for cos⁡(x−1)\cos(x-1) centered at c=1c=1 is obtained by substituting (x−1)(x-1) for xx in the Maclaurin series for cos⁡(x)\cos(x). This gives ∑n=0∞(−1)n(x−1)2n(2n)!\sum_{n=0}^\infty \frac{(-1)^n (x-1)^{2n}}{(2n)!}. Since the original series for cos⁡(u)\cos(u) converges for all real numbers uu, the substituted series will converge for all values of x−1x-1, which means it converges for all real numbers xx. The substitution simply shifts the center of the series expansion; it does not change the radius of convergence. Therefore, the interval of convergence is (−∞,∞)(-\infty, \infty).

Question 12

Suppose the series ∑n=0∞anxn\sum_{n=0}^\infty a_n x^n has radius of convergence Ra=3R_a = 3, and the series ∑n=0∞bnxn\sum_{n=0}^\infty b_n x^n has radius of convergence Rb=7R_b = 7. If cn=an+bnc_n = a_n + b_n, what is the radius of convergence RcR_c of the series ∑n=0∞cnxn\sum_{n=0}^\infty c_n x^n?

  1. Rc=3R_c = 3 (correct answer)
  2. Rc=7R_c = 7
  3. Rc=10R_c = 10
  4. RcR_c cannot be determined from the information given.
Explanation: A power series for a sum of functions converges on the intersection of the individual intervals of convergence. The interval for the first series is (−3,3)(-3,3) (plus maybe endpoints), and for the second is (−7,7)(-7,7) (plus maybe endpoints). The intersection is (−3,3)(-3,3). The radius of convergence for the sum series is the minimum of the individual radii, provided they are not equal. Since Ra=3≠Rb=7R_a = 3 \neq R_b = 7, the radius of convergence of the sum is Rc=min⁡(3,7)=3R_c = \min(3, 7) = 3.

Question 13

Consider the power series ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n where $$a_n = \begin{cases} \frac{1}{2^n} & \text{if } n \text{ is even} \ \frac{1}{3^n} & \text{if } n \text{ is odd} \end{cases}

  1. R=62R = \frac{\sqrt{6}}{2}
  2. R=3R = 3
  3. R=6R = \sqrt{6}
  4. R=2R = 2 (correct answer)
Explanation: When you encounter a power series with coefficients that follow different patterns for even and odd terms, you need to use the root test (Cauchy-Hadamard theorem) rather than the ratio test, since the coefficients don't follow a single pattern. For the radius of convergence, we use R=1lim sup⁡n→∞∣an∣nR = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}. Since the coefficients alternate between two different formulas, we need to find lim sup⁡n→∞∣an∣n\limsup_{n \to \infty} \sqrt[n]{|a_n|} by examining both subsequences. For even nn: ann=12nn=12\sqrt[n]{a_n} = \sqrt[n]{\frac{1}{2^n}} = \frac{1}{2} For odd nn: ann=13nn=13\sqrt[n]{a_n} = \sqrt[n]{\frac{1}{3^n}} = \frac{1}{3} The lim sup⁡\limsup is the larger of these two values, which is 12\frac{1}{2}. Therefore, R=11/2=2R = \frac{1}{1/2} = 2. Choice A (R=62R = \frac{\sqrt{6}}{2}) likely comes from incorrectly trying to find some average or geometric mean of the two patterns. Choice B (R=3R = 3) results from mistakenly taking the reciprocal of 13\frac{1}{3} and ignoring the even terms. Choice C (R=6R = \sqrt{6}) might arise from incorrectly multiplying the individual radii 2×3=62 \times 3 = 6 and taking the square root. Study tip: When coefficients follow different patterns for even and odd indices, always use the root test and remember that lim sup⁡\limsup takes the larger limiting value. The series converges only where both subsequences would converge individually.

Question 14

For the power series ∑n=1∞(−1)nxnnα\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n^{\alpha}} where α>0\alpha > 0, the interval of convergence changes behavior at a critical value of α\alpha. What is this critical value and how does the interval change?

  1. At α=1\alpha = 1, the interval changes from (−1,1)(-1,1) to [−1,1)[-1,1)
  2. At α=12\alpha = \frac{1}{2}, the interval changes from (−1,1)(-1,1) to (−1,1](-1,1]
  3. At α=1\alpha = 1, the interval changes from (−1,1](-1,1] to [−1,1][-1,1] (correct answer)
  4. At α=2\alpha = 2, the interval changes from [−1,1)[-1,1) to [−1,1][-1,1]
Explanation: When analyzing power series convergence, you need to find the radius of convergence first, then test the endpoints separately. The behavior at endpoints depends critically on the series coefficients. For ∑n=1∞(−1)nxnnα\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n^{\alpha}}, the radius of convergence is R=1R = 1 (using the ratio test), so you must check convergence at x=−1x = -1 and x=1x = 1. At x=1x = 1: The series becomes ∑n=1∞(−1)nnα\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\alpha}}, which is an alternating series. By the alternating series test, this converges for any α>0\alpha > 0 since 1nα\frac{1}{n^{\alpha}} decreases to zero. At x=−1x = -1: The series becomes ∑n=1∞(−1)n(−1)nnα=∑n=1∞1nα\sum_{n=1}^{\infty} \frac{(-1)^n (-1)^n}{n^{\alpha}} = \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}, which is a p-series. This converges only when α>1\alpha > 1. Therefore: For 0<α≤10 < \alpha \leq 1, the interval is (−1,1](-1,1]. For α>1\alpha > 1, the interval is [−1,1][-1,1]. The critical transition occurs at α=1\alpha = 1, making C correct. Choice A incorrectly suggests the interval starts as (−1,1)(-1,1) and that x=1x = 1 is excluded after the transition. Choice B uses the wrong critical value and wrong endpoint behavior. Choice D uses α=2\alpha = 2 as the critical value and incorrectly suggests x=1x = 1 was initially excluded. Study tip: For power series endpoint analysis, always check alternating series behavior at one endpoint and p-series behavior at the other. The critical value α=1\alpha = 1 frequently appears because it's where p-series convergence begins.

Question 15

The power series ∑n=0∞(3x−6)nn+1\sum_{n=0}^{\infty} \frac{(3x-6)^n}{\sqrt{n+1}} has interval of convergence II. Which interval represents II?

  1. [53,73]\left[\frac{5}{3}, \frac{7}{3}\right]
  2. (53,73)\left(\frac{5}{3}, \frac{7}{3}\right)
  3. [53,73)\left[\frac{5}{3}, \frac{7}{3}\right) (correct answer)
  4. (53,73]\left(\frac{5}{3}, \frac{7}{3}\right]
Explanation: Rewrite as ∑n=0∞3n(x−2)nn+1\sum_{n=0}^{\infty} \frac{3^n(x-2)^n}{\sqrt{n+1}}. Using ratio test: lim⁡n→∞∣3n+1n+2∣=3\lim_{n \to \infty} \left|\frac{3\sqrt{n+1}}{\sqrt{n+2}}\right| = 3, so radius is R=13R = \frac{1}{3} around x=2x = 2. Series converges for ∣x−2∣<13|x-2| < \frac{1}{3}, i.e., 53<x<73\frac{5}{3} < x < \frac{7}{3}. At x=53x = \frac{5}{3}: ∑n=0∞(−1)nn+1\sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+1}} converges by alternating series test. At x=73x = \frac{7}{3}: ∑n=0∞1n+1\sum_{n=0}^{\infty} \frac{1}{\sqrt{n+1}} diverges (p-series with p=12<1p = \frac{1}{2} < 1). Therefore I=[53,73)I = \left[\frac{5}{3}, \frac{7}{3}\right).

Question 16

The Maclaurin series for arctan⁡(x)\arctan(x) is given by ∑n=0∞(−1)nx2n+12n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}. Determine its interval of convergence.

  1. (−1,1)(-1, 1)
  2. [−1,1)[-1, 1)
  3. (−1,1](-1, 1]
  4. [−1,1][-1, 1] (correct answer)
Explanation: The series for arctan⁡(x)\arctan(x) can be obtained by integrating the geometric series for 11+t2=∑(−1)nt2n\frac{1}{1+t^2} = \sum (-1)^n t^{2n}, which has a radius of convergence R=1R=1. The radius of convergence of the integrated series is also R=1R=1, so it converges on (−1,1)(-1, 1). We must test the endpoints. At x=1x=1, the series is ∑(−1)n2n+1\sum \frac{(-1)^n}{2n+1}, which converges by the Alternating Series Test. At x=−1x=-1, the series is ∑(−1)n(−1)2n+12n+1=∑(−1)3n+12n+1\sum \frac{(-1)^n (-1)^{2n+1}}{2n+1} = \sum \frac{(-1)^{3n+1}}{2n+1}, which also converges by the Alternating Series Test. Therefore, the interval of convergence is [−1,1][-1, 1].

Question 17

If ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n has radius of convergence R=3R = 3, what is the radius of convergence of ∑n=0∞an(x−1)2n2n\sum_{n=0}^{\infty} a_n \frac{(x-1)^{2n}}{2^n}?

  1. 66
  2. 3\sqrt{3}
  3. 33
  4. 6\sqrt{6} (correct answer)
Explanation: When you encounter power series with modified forms, you need to identify how the transformation affects the radius of convergence. The key is recognizing that the new series can be rewritten in terms of a substitution into the original series. Starting with ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n having radius of convergence R=3R = 3, let's analyze ∑n=0∞an(x−1)2n2n\sum_{n=0}^{\infty} a_n \frac{(x-1)^{2n}}{2^n}. First, rewrite this as ∑n=0∞an((x−1)22)n\sum_{n=0}^{\infty} a_n \left(\frac{(x-1)^2}{2}\right)^n. This is exactly the original series with u=(x−1)22u = \frac{(x-1)^2}{2} substituted for xx. Since the original series converges when ∣x∣<3|x| < 3, the transformed series converges when ∣u∣<3|u| < 3, which means ∣(x−1)22∣<3\left|\frac{(x-1)^2}{2}\right| < 3. Solving this inequality: (x−1)22<3\frac{(x-1)^2}{2} < 3, so (x−1)2<6(x-1)^2 < 6, giving us ∣x−1∣<6|x-1| < \sqrt{6}. Therefore, the radius of convergence is 6\sqrt{6}, which is answer D. Looking at the wrong answers: A) 66 likely comes from forgetting to take the square root when solving (x−1)2<6(x-1)^2 < 6. B) 3\sqrt{3} results from incorrectly handling the factor of 2 in the denominator. C) 33 suggests ignoring the transformation entirely and assuming the radius stays the same. Strategy tip: For transformed power series, always substitute the new variable into the convergence condition of the original series. The algebra of solving the resulting inequality determines the new radius—don't try to memorize transformation rules.

Question 18

Find the radius of convergence for the power series ∑n=0∞(3n)!n!(2n)!xn\sum_{n=0}^{\infty} \frac{(3n)!}{n! (2n)!} x^n

  1. R=4/27R = 4/27 (correct answer)
  2. R=27/4R = 27/4
  3. R=1R = 1
  4. R=0R = 0
Explanation: We apply the Ratio Test. L=lim⁡n→∞∣(3(n+1))!xn+1(n+1)!(2(n+1))!⋅n!(2n)!(3n)!xn∣=∣x∣lim⁡n→∞(3n+3)(3n+2)(3n+1)(n+1)(2n+2)(2n+1)L = \lim_{n \to \infty} \left| \frac{(3(n+1))! x^{n+1}}{(n+1)!(2(n+1))!} \cdot \frac{n!(2n)!}{(3n)!x^n} \right| = |x| \lim_{n \to \infty} \frac{(3n+3)(3n+2)(3n+1)}{(n+1)(2n+2)(2n+1)}. The limit of the rational expression in nn is 33n31⋅2⋅n3=274\frac{3^3 n^3}{1 \cdot 2 \cdot n^3} = \frac{27}{4}. So, L=274∣x∣L = \frac{27}{4}|x|. For convergence, we need L<1L < 1, which means 274∣x∣<1\frac{27}{4}|x| < 1, or ∣x∣<427|x| < \frac{4}{27}. The radius of convergence is R=4/27R = 4/27.

Question 19

Determine the interval of convergence for the power series ∑n=1∞n(x+1)n4n(n2+1)\sum_{n=1}^{\infty} \frac{n(x+1)^n}{4^n(n^2+1)}

  1. (−5,3)(-5, 3)
  2. [−5,3)[-5, 3) (correct answer)
  3. [−5,3][-5, 3]
  4. (−3,5)(-3, 5)
Explanation: Using the Ratio Test, the limit of the ratio of absolute values is L=∣x+1∣4lim⁡n→∞(n+1)(n2+1)n((n+1)2+1)=∣x+1∣4L = \frac{|x+1|}{4} \lim_{n \to \infty} \frac{(n+1)(n^2+1)}{n((n+1)^2+1)} = \frac{|x+1|}{4}. Convergence requires L<1L<1, so ∣x+1∣<4|x+1|<4, giving R=4R=4 and an open interval −5<x<3-5 < x < 3. At x=3x=3, the series is ∑nn2+1\sum \frac{n}{n^2+1}, which diverges by limit comparison with the harmonic series ∑1/n\sum 1/n. At x=−5x=-5, the series is ∑(−1)nnn2+1\sum \frac{(-1)^n n}{n^2+1}, which converges by the Alternating Series Test. Thus, the interval of convergence is [−5,3)[-5, 3).

Question 20

A power series ∑n=0∞an(x−3)n\sum_{n=0}^{\infty} a_n (x-3)^n is known to converge at x=0x=0 and diverge at x=8x=8. Which of the following statements must be true?

  1. The series converges at x=7x=7.
  2. The series diverges at x=−2x=-2.
  3. The series converges at x=5x=5. (correct answer)
  4. The series diverges at x=−1x=-1.
Explanation: The series is centered at c=3c=3. Since it converges at x=0x=0, the distance from the center is ∣0−3∣=3|0-3|=3. This implies the radius of convergence RR is at least 3, so R≥3R \ge 3. Since it diverges at x=8x=8, the distance from the center is ∣8−3∣=5|8-3|=5. This implies the radius of convergence RR is at most 5, so R≤5R \le 5. Therefore, we know 3≤R≤53 \le R \le 5. The series must converge for all xx such that ∣x−3∣<R|x-3| < R. For x=5x=5, the distance is ∣5−3∣=2|5-3|=2. Since 2<3≤R2 < 3 \le R, the series must converge at x=5x=5. The other options are not guaranteed: at x=7x=7 (distance 4), convergence depends on whether R>4R>4; at x=−2x=-2 (distance 5) and x=−1x=-1 (distance 4), convergence is not certain.