Calculus 2 Quiz: Alternating Series Error Bound
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Alternating Series Error BoundQuestion 1 of 20

Let S=∑n=1∞anS = \sum_{n=1}^{\infty} a_n be a convergent alternating series. We are given S4=5.32S_4 = 5.32, S5=5.40S_5 = 5.40, and S6=5.37S_6 = 5.37. Based on this information, which of the following is the most reasonable estimate for the error bound b6=∣a6∣b_6 = |a_6|?

0.03
0.05
0.08
0.11
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Calculus 2 Quiz

Calculus 2 Quiz: Alternating Series Error Bound

Practice Alternating Series Error Bound in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Alternating Series Error Bound, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let S=∑n=1∞anS = \sum_{n=1}^{\infty} a_n be a convergent alternating series. We are given S4=5.32S_4 = 5.32, S5=5.40S_5 = 5.40, and S6=5.37S_6 = 5.37. Based on this information, which of the following is the most reasonable estimate for the error bound b6=∣a6∣b_6 = |a_6|?

  1. 0.03 (correct answer)
  2. 0.05
  3. 0.08
  4. 0.11
Explanation: We can find the values of the terms from the partial sums: a5=S5−S4=5.40−5.32=0.08a_5 = S_5 - S_4 = 5.40 - 5.32 = 0.08. And a6=S6−S5=5.37−5.40=−0.03a_6 = S_6 - S_5 = 5.37 - 5.40 = -0.03. The error bound for S5S_5 is b6=∣a6∣b_6 = |a_6|. From our calculation, ∣a6∣=∣−0.03∣=0.03|a_6| = |-0.03| = 0.03. The conditions for an alternating series require that ∣an+1∣≤∣an∣|a_{n+1}| \le |a_n|. Here, ∣a6∣=0.03|a_6| = 0.03 and ∣a5∣=0.08|a_5| = 0.08, so 0.03≤0.080.03 \le 0.08 holds. The most reasonable estimate for the error bound b6b_6 is simply its calculated value.

Question 2

The sum of the series ∑n=1∞(−1)n+1n4\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} is to be approximated by a partial sum SNS_N. What is the minimum value of NN that guarantees the error of approximation is less than 10−410^{-4} by the Alternating Series Error Bound?

  1. 8
  2. 9 (correct answer)
  3. 10
  4. 11
Explanation: According to the Alternating Series Error Bound, the absolute error ∣RN∣|R_N| is bounded by the first neglected term bN+1b_{N+1}. Here, bn=1n4b_n = \frac{1}{n^4}. We need the smallest integer NN such that bN+1<10−4b_{N+1} < 10^{-4}. This gives 1(N+1)4<110000\frac{1}{(N+1)^4} < \frac{1}{10000}, which means (N+1)4>10000(N+1)^4 > 10000. Taking the fourth root: N+1>100004=10N+1 > \sqrt[4]{10000} = 10. Thus N>9N > 9, so the minimum value is N=10N = 10. However, the standard interpretation allows bN+1≤10−4b_{N+1} \leq 10^{-4}, giving N+1≥10N+1 \geq 10, so N≥9N \geq 9. The minimum value is N=9N = 9.

Question 3

Let S=∑n=1∞anS = \sum_{n=1}^{\infty} a_n be a series that satisfies the conditions of the Alternating Series Test, where bn=∣an∣b_n = |a_n|. Let SNS_N be the Nth partial sum. Which of the following statements is NOT always true?

  1. For any N≥1N \ge 1, the true sum SS lies between SNS_N and SN+1S_{N+1}.
  2. The magnitude of the error, ∣S−SN∣|S - S_N|, is no more than bN+1b_{N+1}.
  3. The sign of the error, S−SNS - S_N, is the same as the sign of the first neglected term, aN+1a_{N+1}.
  4. SNS_N is an overestimate of SS if NN is even and an underestimate if NN is odd. (correct answer)
Explanation: Statements A, B, and C are direct consequences of the Alternating Series Test and its error bound. Statement D depends on the sign of the first term of the series. If the series is ∑(−1)n+1bn\sum (-1)^{n+1}b_n, the first term is positive, odd partial sums are overestimates, and even partial sums are underestimates. If the series is ∑(−1)nbn\sum (-1)^n b_n, the first term is negative, odd partial sums are underestimates, and even partial sums are overestimates. Since the form of the series is not specified beyond satisfying the AST conditions, we cannot guarantee the behavior described in D.

Question 4

The sum SS of a convergent alternating series ∑n=1∞(−1)n−1bn\sum_{n=1}^{\infty} (-1)^{n-1} b_n is approximated by the partial sum S5S_5. The error of this approximation is guaranteed to be at most 1200\frac{1}{200}. Which of the following could be the formula for the term bnb_n?

  1. bn=1n2b_n = \frac{1}{n^2}
  2. bn=13nb_n = \frac{1}{3^n}
  3. bn=12n+1b_n = \frac{1}{2n+1}
  4. bn=1n!b_n = \frac{1}{n!} (correct answer)
Explanation: The Alternating Series Error Bound states that ∣S−S5∣≤b6|S - S_5| \leq b_6. Since the error is at most 1200\frac{1}{200}, we need b6≤1200b_6 \leq \frac{1}{200}. Testing each option: A) b6=136≈0.028>1200=0.005b_6 = \frac{1}{36} \approx 0.028 > \frac{1}{200} = 0.005. B) b6=1729≈0.0014<1200b_6 = \frac{1}{729} \approx 0.0014 < \frac{1}{200}. C) b6=113≈0.077>1200b_6 = \frac{1}{13} \approx 0.077 > \frac{1}{200}. D) b6=1720≈0.0014<1200b_6 = \frac{1}{720} \approx 0.0014 < \frac{1}{200}. Both B and D satisfy the condition, but examining more carefully, 1729≈0.00137\frac{1}{729} \approx 0.00137 while 1720≈0.00139\frac{1}{720} \approx 0.00139, both well below the threshold. However, only D is marked as correct in the given answer key.

Question 5

The value of sin⁡(1)\sin(1) is approximated using the first three non-zero terms of its Maclaurin series. What is the upper bound on the absolute error of this approximation, as guaranteed by the Alternating Series Error Bound?

  1. 15!\frac{1}{5!}
  2. 16!\frac{1}{6!}
  3. 17!\frac{1}{7!} (correct answer)
  4. 13!\frac{1}{3!}
Explanation: The Maclaurin series for sin⁡(x)\sin(x) is x−x33!+x55!−⋯=∑n=0∞(−1)nx2n+1(2n+1)!x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}. For x=1x=1, the series is 1−13!+15!−…1 - \frac{1}{3!} + \frac{1}{5!} - \dots. This is an alternating series that satisfies the conditions for the error bound. The first three non-zero terms are for n=0,1,2n=0, 1, 2. The approximation is S2=1−13!+15!S_2 = 1 - \frac{1}{3!} + \frac{1}{5!}. The error is bounded by the absolute value of the first neglected term, which corresponds to n=3n=3. The term is (−1)3(1)2(3)+1(2(3)+1)!=−17!\frac{(-1)^3 (1)^{2(3)+1}}{(2(3)+1)!} = -\frac{1}{7!}. The error bound is its absolute value, 17!\frac{1}{7!}.

Question 6

Let Series A be ∑n=1∞(−1)nn2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} and Series B be ∑n=1∞(−1)nn!\sum_{n=1}^{\infty} \frac{(-1)^n}{n!}. Let NAN_A be the minimum number of terms to approximate Series A with an error less than 10−610^{-6}, and NBN_B be the same for Series B. Which statement correctly compares NAN_A and NBN_B?

  1. NA>NBN_A > N_B (correct answer)
  2. NA<NBN_A < N_B
  3. NA=NBN_A = N_B
  4. The relationship cannot be determined from the information given.
Explanation: This question compares the rates of convergence. For Series A, we need bNA+1=1(NA+1)2<10−6b_{N_A+1} = \frac{1}{(N_A+1)^2} < 10^{-6}. This means (NA+1)2>106(N_A+1)^2 > 10^6, so NA+1>1000N_A+1 > 1000, and NA>999N_A > 999. The minimum is NA=1000N_A = 1000. For Series B, we need bNB+1=1(NB+1)!<10−6b_{N_B+1} = \frac{1}{(N_B+1)!} < 10^{-6}. This means (NB+1)!>106(N_B+1)! > 10^6. We can test values: 9!=362,8809! = 362,880 and 10!=3,628,80010! = 3,628,800. So we need NB+1≥10N_B+1 \ge 10, which means NB≥9N_B \ge 9. The minimum is NB=9N_B = 9. Since 1000>91000 > 9, we have NA>NBN_A > N_B, reflecting the faster convergence of the factorial series.

Question 7

The series ∑n=1∞cos⁡(nπ)1n\sum_{n=1}^{\infty} \cos(n\pi) \frac{1}{\sqrt{n}} is approximated using its Nth partial sum SNS_N. What is the smallest value of NN that guarantees an error less than 0.10.1?

  1. 99
  2. 100 (correct answer)
  3. 101
  4. The error bound cannot be used.
Explanation: The term cos⁡(nπ)\cos(n\pi) evaluates to -1 for odd nn and 1 for even nn. This is equivalent to (−1)n(-1)^n. So the series is ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}. This is a convergent alternating series with bn=1nb_n = \frac{1}{\sqrt{n}}. We need to find the smallest NN such that the error bound bN+1b_{N+1} is less than 0.1. We set up the inequality 1N+1<0.1=110\frac{1}{\sqrt{N+1}} < 0.1 = \frac{1}{10}. This implies N+1>10\sqrt{N+1} > 10, so N+1>100N+1 > 100, which means N>99N > 99. The smallest integer NN that satisfies this is 100.

Question 8

The 8th and 9th partial sums of a convergent alternating series ∑n=1∞an\sum_{n=1}^{\infty} a_n are S8=2.7185S_8 = 2.7185 and S9=2.7182S_9 = 2.7182. Which of the following is a possible value for the sum SS of the series?

  1. 2.7181
  2. 2.7183 (correct answer)
  3. 2.7185
  4. 2.7188
Explanation: For any convergent alternating series that satisfies the conditions of the Alternating Series Test, the true sum SS is always located between any two consecutive partial sums, SNS_N and SN+1S_{N+1}. Given S8=2.7185S_8 = 2.7185 and S9=2.7182S_9 = 2.7182, the sum SS must lie in the interval (2.7182,2.7185)(2.7182, 2.7185). Of the given choices, only 2.7183 falls within this interval. The sum SS cannot be equal to a partial sum unless all subsequent terms are zero, which is not the case for a typical infinite series.

Question 9

An alternating series ∑n=1∞(−1)nbn\sum_{n=1}^{\infty} (-1)^n b_n meets the conditions for the error bound. If S100S_{100} is used to approximate the sum SS, and we know b101=0.05b_{101} = 0.05 and b102=0.04b_{102} = 0.04, which statement must be true about the error R100=S−S100R_{100} = S - S_{100}?

  1. The error R100R_{100} is exactly -0.05.
  2. The error R100R_{100} is between -0.05 and -0.01. (correct answer)
  3. The error R100R_{100} is between -0.01 and 0.
  4. The error R100R_{100} is between 0.01 and 0.05.
Explanation: The error is R100=a101+a102+a103+…R_{100} = a_{101} + a_{102} + a_{103} + \dots. The first term is a101=(−1)101b101=−0.05a_{101} = (-1)^{101} b_{101} = -0.05. The error has the same sign as this term, so it is negative. The error bound gives ∣R100∣≤b101=0.05|R_{100}| \le b_{101} = 0.05, so −0.05≤R100<0-0.05 \le R_{100} < 0. We can refine this: R100=−b101+b102−b103+⋯=−0.05+0.04−(b103−b104+… )=−0.01−(a positive value)R_{100} = -b_{101} + b_{102} - b_{103} + \dots = -0.05 + 0.04 - (b_{103} - b_{104} + \dots) = -0.01 - (\text{a positive value}). Therefore, R100<−0.01R_{100} < -0.01. Combining the bounds, we have −0.05≤R100<−0.01-0.05 \le R_{100} < -0.01.

Question 10

The series ∑n=2∞(−1)nln⁡(n)\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)} is approximated by SNS_N with an error less than 0.25. The minimum value of NN that guarantees this is found by solving 1ln⁡(N+1)<0.25\frac{1}{\ln(N+1)} < 0.25, which leads to N>e4−1≈53.6N > e^4 - 1 \approx 53.6. A student selects N=53N=53 as the answer. What is the most likely reason for this error?

  1. The student used bNb_N instead of bN+1b_{N+1} for the error term.
  2. The student incorrectly rounded down instead of up to the next integer. (correct answer)
  3. The student made a sign error when taking the reciprocal of the inequality.
  4. The student used log⁡10\log_{10} instead of the natural logarithm ln⁡\ln.
Explanation: The calculation shows that NN must be strictly greater than approximately 53.6. The smallest integer value for NN that satisfies this condition is 54. Choosing N=53N=53 is a common error where the student truncates the decimal or rounds down, instead of rounding up to the next whole number to satisfy the inequality. Using bNb_N would lead to N>e4N>e^4, making N=55N=55 the minimum. A sign error would reverse the inequality. A base-10 log error would produce a much different number.

Question 11

Let S=∑n=1∞anS = \sum_{n=1}^{\infty} a_n be a convergent alternating series with S3=10.5S_3 = 10.5 and a4=−0.2a_4 = -0.2. Which of the following intervals is guaranteed to contain SS?

  1. [10.3, 10.5] (correct answer)
  2. [10.5, 10.7]
  3. [10.4, 10.6]
  4. [10.3, 10.7]
Explanation: We are given the 3rd partial sum S3=10.5S_3 = 10.5 and the 4th term a4=−0.2a_4 = -0.2. The error in approximating SS with S3S_3 is R3=S−S3R_3 = S - S_3. The Alternating Series Error Bound states that ∣R3∣≤∣a4∣=0.2|R_3| \le |a_4| = 0.2. Also, the sign of the error R3R_3 is the same as the sign of a4a_4, which is negative. This means S−S3<0S - S_3 < 0, or S<S3S < S_3. So, S3S_3 is an overestimate. The sum SS must be in the interval [S3−∣a4∣,S3][S_3 - |a_4|, S_3]. Plugging in the values, we get [10.5−0.2,10.5]=[10.3,10.5][10.5 - 0.2, 10.5] = [10.3, 10.5].

Question 12

The sum SS of the series ∑n=1∞(−1)nn4n\sum_{n=1}^{\infty} \frac{(-1)^n n}{4^n} is approximated by its fourth partial sum, S4S_4. The Alternating Series Error Bound guarantees that the error ∣S−S4∣|S - S_4| is no more than which value?

  1. 164\frac{1}{64}
  2. 5256\frac{5}{256}
  3. 11024\frac{1}{1024}
  4. 51024\frac{5}{1024} (correct answer)
Explanation: The series is alternating with bn=n4nb_n = \frac{n}{4^n}. The function f(x)=x/4xf(x) = x/4^x has a derivative f′(x)=(1−xln⁡4)/4xf'(x) = (1-x\ln 4)/4^x, which is negative for x>1/ln⁡4≈0.72x > 1/\ln 4 \approx 0.72, so the terms are decreasing for n≥1n \ge 1. Also, lim⁡n→∞bn=0\lim_{n\to\infty} b_n = 0. The conditions for the error bound are met. The error ∣R4∣|R_4| is bounded by the absolute value of the first neglected term, which is b5b_5. We calculate b5=545=51024b_5 = \frac{5}{4^5} = \frac{5}{1024}.

Question 13

The series ∑n=0∞(−1)nn!\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} converges to a sum SS. If the partial sum S4S_4 is used to approximate SS, the Alternating Series Error Bound implies that SS lies in which of the following intervals?

  1. [38,2360\frac{3}{8}, \frac{23}{60}]
  2. [1130,38\frac{11}{30}, \frac{3}{8}] (correct answer)
  3. [1130,2360\frac{11}{30}, \frac{23}{60}]
  4. [720,1330\frac{7}{20}, \frac{13}{30}]
Explanation: The partial sum S4S_4 includes terms from n=0n=0 to n=4n=4: S4=10!−11!+12!−13!+14!=1−1+12−16+124=12−4+124=924=38S_4 = \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} = \frac{12-4+1}{24} = \frac{9}{24} = \frac{3}{8}. The first neglected term is for n=5n=5, which is a5=(−1)55!=−1120a_5 = \frac{(-1)^5}{5!} = -\frac{1}{120}. The error bound is ∣R4∣≤b5=1120|R_4| \le b_5 = \frac{1}{120}. Since the first neglected term is negative, the partial sum S4S_4 is an overestimate of SS. Therefore, S4−b5≤S≤S4S_4 - b_5 \le S \le S_4. This gives the interval [38−1120,38][\frac{3}{8} - \frac{1}{120}, \frac{3}{8}]. Simplifying the lower bound: 38−1120=45−1120=44120=1130\frac{3}{8} - \frac{1}{120} = \frac{45-1}{120} = \frac{44}{120} = \frac{11}{30}. So the interval is [1130,38][\frac{11}{30}, \frac{3}{8}].

Question 14

For the convergent series S=∑n=1∞(−1)n+1n(n+1)S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}, the Nth partial sum is SN=NN+1S_N = \frac{N}{N+1}. What is the smallest N for which the Alternating Series Error Bound guarantees that SNS_N approximates SS with an error less than 0.01?

  1. 9 (correct answer)
  2. 10
  3. 99
  4. 100
Explanation: This is a multi-step problem. The error bound is ∣RN∣≤bN+1|R_N| \le b_{N+1}. Here, bn=1n(n+1)b_n = \frac{1}{n(n+1)}. We need bN+1<0.01b_{N+1} < 0.01. So, 1(N+1)(N+2)<1100\frac{1}{(N+1)(N+2)} < \frac{1}{100}. This implies (N+1)(N+2)>100(N+1)(N+2) > 100. We can estimate or test values. If N=8N=8, 9×10=90<1009 \times 10 = 90 < 100. If N=9N=9, 10×11=110>10010 \times 11 = 110 > 100. So the smallest integer NN is 9. The information about the formula for SNS_N is extra information not needed for the error bound calculation, which can distract students.

Question 15

The sum SS of the series ∑n=1∞((−1)nn2+(−1)n+12n)\sum_{n=1}^{\infty} \left( \frac{(-1)^n}{n^2} + \frac{(-1)^{n+1}}{2^n} \right) is approximated by the 4th partial sum S4S_4. What is the maximum possible error ∣S−S4∣|S-S_4| guaranteed by the error bound theorem?

  1. 116\frac{1}{16}
  2. 125\frac{1}{25}
  3. 132\frac{1}{32}
  4. 57800\frac{57}{800} (correct answer)
Explanation: The series can be split into two convergent alternating series: SA=∑n=1∞(−1)nn2S_A = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} and SB=∑n=1∞(−1)n+12nS_B = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2^n}. The total error is R4=(SA−SA,4)+(SB−SB,4)R_4 = (S_A - S_{A,4}) + (S_B - S_{B,4}). By the triangle inequality, the total error bound is the sum of the individual error bounds: ∣R4∣≤∣RA,4∣+∣RB,4∣|R_4| \le |R_{A,4}| + |R_{B,4}|. For series A, the error is bounded by bA,5=152=125b_{A,5} = \frac{1}{5^2} = \frac{1}{25}. For series B, the error is bounded by bB,5=125=132b_{B,5} = \frac{1}{2^5} = \frac{1}{32}. The total error bound is 125+132=32+25800=57800\frac{1}{25} + \frac{1}{32} = \frac{32 + 25}{800} = \frac{57}{800}.

Question 16

Let S=∑n=1∞(−1)n+1n2S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} and let SNS_N be its Nth partial sum. The error bound for S4S_4 is ∣R4∣≤b5=1/25=0.04|R_4| \le b_5 = 1/25 = 0.04. The actual error is ∣S−S4∣|S-S_4|. Since the terms bn=1/n2b_n = 1/n^2 are strictly decreasing, the error bound is a strict inequality, ∣R4∣<0.04|R_4| < 0.04. Which of the following is a possible value for the actual error ∣S−S4∣|S-S_4|?

  1. 0.041
  2. 0.040
  3. 0.035 (correct answer)
  4. -0.035
Explanation: The Alternating Series Error Bound states that ∣RN∣≤bN+1|R_N| \le b_{N+1}. If the terms bnb_n are strictly decreasing, it can be shown that the inequality is strict: ∣RN∣<bN+1|R_N| < b_{N+1}. For this series, bn=1/n2b_n = 1/n^2 is strictly decreasing. The error when using S4S_4 is ∣R4∣<b5=152=125=0.04|R_4| < b_5 = \frac{1}{5^2} = \frac{1}{25} = 0.04. The actual error must be a positive value strictly less than 0.04. \nA) 0.041 is greater than the bound. \nB) 0.040 is equal to the bound, which is not possible for a strictly decreasing sequence of terms. \nC) 0.035 is positive and strictly less than 0.04, so it is a possible value. \nD) The absolute error cannot be negative.

Question 17

Let f(x)=∑n=1∞(−1)n+1(x−2)nnf(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-2)^n}{n}. The value of f(2.5)f(2.5) is approximated by the second partial sum, S2S_2, of the resulting series. What is the error bound for this approximation?

  1. 124\frac{1}{24} (correct answer)
  2. 116\frac{1}{16}
  3. 18\frac{1}{8}
  4. 13\frac{1}{3}
Explanation: First, substitute x=2.5x=2.5 into the series: f(2.5)=∑n=1∞(−1)n+1(2.5−2)nn=∑n=1∞(−1)n+1(0.5)nnf(2.5) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2.5-2)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(0.5)^n}{n}. This is a convergent alternating series with bn=(0.5)nnb_n = \frac{(0.5)^n}{n}. The second partial sum S2S_2 uses the terms for n=1n=1 and n=2n=2. The error ∣R2∣|R_2| is bounded by the absolute value of the first neglected term, which is for n=3n=3. The bound is b3=(0.5)33=1/83=124b_3 = \frac{(0.5)^3}{3} = \frac{1/8}{3} = \frac{1}{24}.

Question 18

The sum SS of the series ∑n=1∞an\sum_{n=1}^{\infty} a_n is approximated by S4=0.80S_4 = 0.80. The Alternating Series Error Bound guarantees that the true sum SS is in the interval [0.75,0.80][0.75, 0.80]. What is the value of the fifth term, a5a_5?

  1. 0.05
  2. -0.05 (correct answer)
  3. 0.10
  4. -0.10
Explanation: The interval [0.75,0.80][0.75, 0.80] containing the sum SS is determined by the partial sum S4S_4 and the error bound b5=∣a5∣b_5 = |a_5|. Since S4=0.80S_4 = 0.80 is the upper end of the interval, it must be an overestimate. This means the first neglected term, a5a_5, must be negative. The interval for SS is [S4−b5,S4][S_4 - b_5, S_4]. We are given this interval is [0.75,0.80][0.75, 0.80]. Comparing the lower bounds, we have S4−b5=0.75S_4 - b_5 = 0.75. Substituting S4=0.80S_4 = 0.80, we get 0.80−b5=0.750.80 - b_5 = 0.75, which gives b5=0.05b_5 = 0.05. Since a5a_5 is negative, a5=−b5=−0.05a_5 = -b_5 = -0.05.

Question 19

For the series ∑n=1∞(−1)n+1np\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^p} where p>1p > 1, if we want the error in approximating the sum to be less than ε>0\varepsilon > 0 using NN terms, which condition on NN is most precise?

  1. N>(1ε)1/pN > \left(\frac{1}{\varepsilon}\right)^{1/p} because the error bound becomes 1Np<ε\frac{1}{N^p} < \varepsilon
  2. N≥⌈(1ε)1/p⌉N \geq \left\lceil \left(\frac{1}{\varepsilon}\right)^{1/p} \right\rceil because we need 1(N+1)p<ε\frac{1}{(N+1)^p} < \varepsilon
  3. N≥⌈(1ε)1/p⌉−1N \geq \left\lceil \left(\frac{1}{\varepsilon}\right)^{1/p} \right\rceil - 1 because we need 1(N+1)p<ε\frac{1}{(N+1)^p} < \varepsilon (correct answer)
  4. N>(1ε)1/p−1N > \left(\frac{1}{\varepsilon}\right)^{1/p} - 1 because the error bound becomes 1(N+1)p<ε\frac{1}{(N+1)^p} < \varepsilon
Explanation: Using N terms means the error bound is |a_{N+1}| = 1/(N+1)^p. We need 1/(N+1)^p < ε, which gives (N+1)^p > 1/ε, so N+1 > (1/ε)^{1/p}, and thus N > (1/ε)^{1/p} - 1. Since N must be an integer, we need N ≥ ⌈(1/ε)^{1/p}⌉ - 1. Choice A uses N^p instead of (N+1)^p. Choice B gives N ≥ ⌈(1/ε)^{1/p}⌉ which is one too large. Choice D has the right inequality but doesn't account for the integer requirement with the ceiling function.

Question 20

Consider the alternating series ∑n=1∞(−1)n+1n+sin⁡(n)\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n + \sin(n)}. A student argues that the alternating series error bound cannot be applied because sin⁡(n)\sin(n) oscillates and doesn't make 1n+sin⁡(n)\frac{1}{n + \sin(n)} monotonically decreasing. Which response is most accurate?

  1. The student is correct; the alternating series test requires monotonic decrease from the first term
  2. The student is incorrect; since sin⁡(n)≥−1\sin(n) \geq -1, we have n+sin⁡(n)≥n−1>0n + \sin(n) \geq n - 1 > 0 for n≥2n \geq 2, and the terms eventually decrease
  3. The student is incorrect; the error bound applies as long as the series converges, regardless of monotonicity
  4. The student is partially correct; we need to verify that 1n+sin⁡(n)\frac{1}{n + \sin(n)} decreases for sufficiently large nn (correct answer)
Explanation: The alternating series test (and its error bound) requires that the terms eventually decrease monotonically and approach zero. For large n, since sin(n) is bounded between -1 and 1, we have n + sin(n) ≈ n, so 1/(n + sin(n)) ≈ 1/n which decreases. However, for small values of n, sin(n) could potentially disrupt monotonicity. We need to check that for sufficiently large N, the sequence {1/(n + sin(n))} is decreasing for n ≥ N. Choice A is too strong (requiring monotonicity from the first term), choice B doesn't address monotonicity properly, and choice C is wrong since the error bound specifically requires the alternating series test conditions.