MATH 1C Final Exam: Sequences and Series Answer Key

MATH 1C: FINAL EXAM, V0A c Jeffrey A. Anderson ANSWER KEY True/False For the problems below, circle T if the answer is true and circle F is the answer is false. After you’ve chosen your answer, mark the appropriate space on your Scantron form. Notice that letter A corresponds to true while letter B corresponds to false. 1. T F ∞ ∑ n =0 ( - 1) n n ! = 1 e 2. T F If ∞ ∑ n =0 a n is divergent, then ∞ ∑ n =0 | a n | is divergent. 3. T F If lim n →∞ a n = 0, then ∞ ∑ n =1 a n converges. 4. T F If a n > 0 and ∞ ∑ n =1 a n converges, then ∞ ∑ n =1 ( - 1) n a n converges. 5. T F If 0 ≤ a n ≤ b n for all n and ∞ ∑ n =0 b n diverges, then ∞ ∑ n =0 a n diverges. 6. T F If a n > 0 for all n ∈ N and lim n →∞ a n +1 a n < 1, then lim n →∞ a n = 0. 7. T F The ratio test can be used to determine if the series ∞ ∑ n =0 1 n 3 converges 8. T F The series ∞ ∑ n =1 3 ne - n 2 diverges

Multiple Choice For the problems below, circle the correct response for each question. After you’ve chosen your answer, mark your answer on your Scantron form. 9. For which of the following values of k do both ∞ ∑ n =0 3 k n and ∞ ∑ n =0 (3 - k ) n √ n + 3 converge? A. None B. 2 C. 3 D. 4 E. 5 10. The series ∞ ∑ n =1 1 n α converges if and only if A. 1 < α B. α < 1 C. - 1 < α < 1 D. α ≥ 1 E. - 1 < α 11. Which of the following series converge? I. ∞ X n =2 1 n (ln( n )) 4 II. ∞ X n =1 3 + sin( n ) n 4 III. ∞ X n =1 7 n 2 - 5 e n ( n + 3) 2 A. I only B. I and II only C. II and III only D. I and III only E. I, II, and III 12. Which of the following series converge? I. ∞ X n =2 1 n 2 · ln( n ) II. ∞ X n =2 1 n · ln( n ) III. ∞ X n =2 1 √ n · ln( n ) A. I only B. II only C. I and II only D. I and III only E. I, II, and III 13. Which of the following is the radius of convergence for the series ∞ ∑ n =0 ( - 1) n n ! x n n n ? A. 0 B. 1 e C. 1 D. e E. ∞ M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 2 of 12

14. What are the values of x for which the series ( x + 1) - ( x + 1) 2 2 + ( x + 1) 3 3 - · · · + ( - 1) n +1 ( x + 1) n n + · · · converges? A. - 2 < x ≤ 0 B. All real numbers C. - 2 < x < 0 D. - 2 ≤ x < 0 E. - 2 ≤ x ≤ 0 15. For what real values of k could the series ∞ X n =1 ln( n ) + n 3 n 5 k + 4 converges? A. k ≥ 1 B. k < 1 C. all real k D. k > 1 / 5 E. k > 4 / 5 16. Find the values of x for which the series ∞ ∑ n =1 ( x - 1) n : A. - 2 < x < 0 B. 0 < x ≤ 2 C. 0 < x < 2 D. 0 ≤ x ≤ 2 E. - 2 ≤ x < 0 17. What are all values of x for which the series ∞ X n =1 (2 x - 1) n 3 √ n 2 converge? A. - 1 ≤ x ≤ 1 B. 0 < x < 1 C. 0 ≤ x < 1 D. 0 < x ≤ 1 E. - 1 ≤ x ≤ 1 18. The series ∞ ∑ n =0 r n converges if and only if: A. - 1 < r < 1 B. - 1 ≤ r ≤ 1 C. - 1 ≤ r < 1 D. - 1 < r ≤ 1 E. r < 1 M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 3 of 12

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19. Consider the series 1 2 + ( x - 25) 4 + ( x - 25) 2 8 + · · · + ( x - 25) n 2 n +1 + · · · Which of the following statements are true about this series? I. The interval where the series converges is 23 < x < 27 II. The sum of the series where it converges is equal to 1 27 - x III. If x = 28 , the series converges to 1 3 IV. The interval where the series converges is 21 < x < 29 V. The series diverges A. I only B. III only C. I and II only D. I and IV only E. V only 20. How many terms of the alternating series ∞ X n =1 ( - 1) n +1 n - 2 must we add in order to be sure that the partial sum s n is within 0 . 0001 of the sum s . A. 10 B. 300 C. 30 D. 100 E. 1000 21. The convergent power series in x that has the same equal to x 3 2 - x 3 when k x k < 3 √ 2 is: A. 1 2 ∞ ∑ n =0 x 3 n +3 2 n B. ∞ ∑ n =0 x 3 n +3 2 n C. ∞ ∑ n =0 x 3 n - 3 4 n D. 1 2 ∞ ∑ n =0 x 3 n 2 n E. 1 2 ∞ ∑ n =0 x 3 n +3 4 n 22. Which of the following is the power series representation of x e 2 x +1 ? A. x - 2 · x 2 + 2 2 · x 3 2! - 2 3 · x 4 3! + · · · B. e · x 2 + 2 e · x 4 + 2 2 e · x 6 2! + 2 3 e · x 8 3! + · · · C. e · x + 2 e · x 2 + 2 2 e · x 3 2! + 2 3 e · x 4 3! + · · · D. x + (2 x + 1) x + (2 x + 1) 2 2! + (2 x + 1) 3 3! + · · · E. x 2 + x 4 + x 6 2! + x 8 3! + · · · M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 4 of 12

23. Which of the following series converge? 1) ∞ X n =1 1 n 2) ∞ X n =1 ( - 1) n · n ln( n ) 3) ∞ X n =1 ( - 1) n n A. 1 B. 2 C. 3 D. 2 , 3 E. None 24. Which of the following is a power series representation for f ( x ) = 1 4 + 6 x if | x | < 2 3 ? A. 1 4   ∞ ∑ n =0 ( - 1) n 3 2 n x n   B. 1 6   ∞ ∑ n =0 ( - 1) n 2 3 n x n   C. 1 4   ∞ ∑ n =0 ( - 1) 2 n 3 4 n x n   D. 1 4   ∞ ∑ n =0 3 2 n x n   E. 1 6   ∞ ∑ n =0 ( - 1) n 3 2 n x n   25. Which of the following is a power series representation for f ( x ) = x · (arctan( x )) if | x | < 1? A. x + x 3 3 + x 5 5 + x 7 7 + x 9 9 + · · · B. 1 - x + x 3 3 - x 5 5 + x 7 7 - x 9 9 + · · · C. 1 + x + x 3 3 + x 5 5 + x 7 7 + x 9 9 + x 11 11 + · · · D. x 2 - x 4 3 + x 6 5 - x 8 7 + x 10 9 + · · · E. x - x 3 3 + x 5 5 - x 7 7 + x 9 9 + · · · M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 5 of 12

26. Which of the following is the power series representation of ln(1 + 2 x )? A. 2 x - 4 x 2 2 + 8 x 3 3 - 16 x 4 4 + 32 x 5 5 + · · · B. x 3 3 - 4 x 4 4 + 8 x 5 5 - 16 x 6 6 + 32 x 7 7 + · · · C. 2 - 4 x + 8 x 2 - 16 x 3 + 32 x 4 + · · · D. 2 x 2 - 4 x 4 + 8 x 6 - 16 x 8 + 32 x 10 + · · · E. 2 x 2 - 4 x 4 2 + 8 x 6 3 - 16 x 8 4 + 32 x 10 5 + · · · 27. Which of the following is a power series representation for f ( x ) = 3 (1 + x ) 2 if | x | < 1? A. 2 x 2 - 3 x 3 + 4 x 4 - 5 x 5 + · · · B. 3 - 3 x + 3 x 2 - 3 x 3 + · · · C. 1 - x + x 2 - x 3 + · · · D. 3 - 3 · (2 x ) + 3 · (3 x 2 ) - 3 · (4 x 3 ) + · · · E. 1 - 2 x + 3 x 2 - 4 x 3 + · · · 28. Which of the following series converge? 1) ∞ X n =1 3 2 n 2 3 n 2) ∞ X n =1 1 ( n + 1) 3 3) ∞ X n =1 n + 1 √ n 3 + 2 A. None B. 1 C. 2 D. 3 E. 2 , 3 29. The Taylor Series about x = 2 for a certain function f converges to f ( x ) for all x in the radius of convergence. The n th derivative of f at x = 2 is given by f ( n ) ( x ) = 2 n n ! 3 n +1 ( n + 1) and f (2) = 3 . Which of the following is the radius of convergence for the Taylor series for f about x = 2? A. - 1 2 B. 2 3 C. 0 D. 3 2 E. 3 M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 6 of 12

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30. Let f be a function having derivatives of all orders for all real numbers x . The third-order Taylor polynomial for f about x = 4 is given by T ( x ) = 1 9 + 5( x - 4) 2 - 8( x - 4) 3 . If f (4) ( x ) ≤ 1 4 for 3 . 5 ≤ x ≤ 4, order the following from greatest to least? I. The maximum value of | T ( x ) - f ( x ) | for 3 . 5 ≤ x ≤ 4 II. | f 00 (4) | III. | f (4) | A. I, II, III B. II, I, III C. III, I, II D. II, III, I E. III, II, I 31. Suppose the second-degree Taylor polynomial for f ( x ) around x = 0 is given by P ( x ) = 7 x + 5 2 x 2 If g ( x ) is the inverse of f ( x ) , what is g 0 (0)? A. 1 B. 1 5 C. 1 7 D. 5 E. Undefined 32. Find the limit of the sequence a n = 2 + - 4 5 n : A. 2 B. 6 5 C. - 4 5 D. - 2 E. 4 5 33. The coefficient of ( x - 3) 2 in the Taylor series for f ( x ) = arctan( x ) about x = 3 is: A. 3 100 B. - 3 100 C. - 3 50 D. 3 50 E. 1 256 M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 7 of 12

Free Response 34. (10 points) Find the value of ∞ ∑ n =2 3 n + 5 n 15 n . Show your work. Solution: ∞ X n =2 3 n + 5 n 15 n = ∞ X n =2 3 n 15 n + 5 n 15 n = ∞ X n =2 1 5 n + 1 3 n = ∞ X n =1 1 5 n +1 + 1 3 n +1 = ∞ X n =1 1 5 2 · 1 5 n - 1 + 1 3 2 · 1 3 n - 1 = 1 25 · 1 1 - 1 5 + 1 9 · 1 1 - 1 3 = 13 60 M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 8 of 12

35. (10 points) Show that the series ∞ ∑ n =2 1 n · (ln( n )) p converges if p > 1 and diverges if p ≤ 1. Solution: In this problem, we are asked to find values of p such that our given series converges. To this end, let f ( x ) = 1 x · (ln( x )) p . For x ≥ 2, we confirm that f ( x ) is positive, decreasing and continuous using methods from Math 1A. Thus, by the integral test for convergence, we know that our given series and the integral ∞ Z 2 f ( x ) dx have identical convergence behavior (the either both converge or they both diverge). With this in mind, consider the two cases: I. Case p = 1: ∞ Z 2 f ( x ) dx = ∞ Z 2 1 x · ln( x ) dx = ∞ Z ln(2) 1 u du Let u = ln( x ) = ⇒ du = 1 x dx and u (2) = ln(2) while u ( ∞ ) = ∞ = lim t →∞ t Z ln(2) 1 u du = lim t →∞ ln( u ) t ln(2) = lim t →∞ ln( t ) - ln(2) = ∞ . Thus, for p = 1 our integral diverges. By the integral test for series, we conclude that the given series also diverges for p = 1. M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 9 of 12

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II. Case p 6 = 1: ∞ Z 2 f ( x ) dx = ∞ Z 2 1 x · (ln( x )) p dx = ∞ Z ln(2) 1 u p du Let u = ln( x ) = ⇒ du = 1 x dx and u (2) = ln(2) while u ( ∞ ) = ∞ = lim t →∞ t Z ln(2) u - p du = lim t →∞ u 1 - p 1 - p t ln(2) =   lim t →∞ t 1 - p 1 - p   + c where c = - (ln(2)) 1 - p 1 - p . Thus we see that the integral converges if p > 1 and diverges for all p ≤ 1. This is what was to be shown. M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 10 of 12

36. (10 points) Use a Taylor Polynomial to estimate the value of the integral 0 . 5 Z 0 ln(1 + x 2 ) dx with an absolute error of at most 1 / 1000. Justify your answer. Solution: Recall from lesson 21 that the power series representation of the natural log function is given by ln(1 + u ) = u - u 2 2 + u 3 3 - u 4 4 + · · · = ∞ X n =1 ( - 1) n - 1 u n n By Theorem 9.4.3 on page 679 concerning Combining Power Series via composition, we know that for u = x 2 , we have ln(1 + x 2 ) = x 2 - x 4 2 + x 6 3 - x 8 4 + · · · = ∞ X n =1 ( - 1) n - 1 x 2 n n In this problem, we are asked to find a definite integral over the interval [0 , 0 . 5] involving the function ln(1+ x 2 ). We know by Theorem 9.5.2 on page 680, that when integrating the power series representation of this function, we can integrate the series term by term. Mathematically, we have 0 . 5 Z 0 ln(1 + x 2 ) dx = 0 . 5 Z 0 " ∞ X n =1 ( - 1) n - 1 x 2 n n # dx = ∞ X n =1   0 . 5 Z 0 ( - 1) n - 1 x 2 n n   dx = ∞ X n =1   ( - 1) n - 1 · 1 n · x 2 n +1 2 n + 1 0 . 5 0   = ∞ X n =1 ( - 1) n - 1 · 1 n · 1 2 n + 1 · 1 2 2 n +1 = ( 0 . 5 ) 3 3 - ( 0 . 5 ) 5 10 + ( 0 . 5 ) 7 21 - ( 0 . 5 ) 9 36 + ( 0 . 5 ) 11 55 - ( 0 . 5 ) 13 78 + · · · This is an alternating series and we can apply The Remainder Theorem 8.20 on page 652. Specifically, we see that since the third term ( 0 . 5 ) 7 21 ≈ 0 . 000372 < 0 . 001 = 1 1000 we know we can add the first to terms to approximate our integral to the desired accuracy. Calcu- lating this approximation, we see 0 . 5 Z 0 ln(1 + x 2 ) dx ≈ ( 0 . 5 ) 3 3 - ( 0 . 5 ) 5 10 = 0 . 0385416 6 . M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 11 of 12

Challenge Problem 37. (Optional, Extra Credit, Challenge Problem) Suppose that a sequence { a n } ∞ n =1 satisfies 0 < a n ≤ a 2 n + a 2 n +1 for all n ≥ 1. Prove that the series ∞ ∑ n =1 a n diverges. Solution: Let us begin this problem by considering the significance of the stated property that 0 < a n ≤ a 2 n + a 2 n +1 for our sequence { a n } ∞ n =1 . Consider a 1 ≤ a 2 + a 3 a 2 + a 3 ≤ a 4 + a 5 + a 6 + a 7 a 4 + a 5 + a 6 + a 7 ≤ a 8 + a 9 + a 10 + a 11 + a 12 + a 13 + a 14 + a 15 Notice we can find a pattern in the indices for these inequalities: 1 ≤ 2 + 3 2 + 3 ≤ 4 + 5 + 6 + 7 4 + 5 + 6 + 7 ≤ 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 Let’s use this pattern to generalize to the k th inequality. Set p k = 2 k - 1 X j =2 k - 1 a j and notice we can write p k ≤ p k +1 . Thus, we must have 2 N - 1 X n =1 a j = N X k =1 p k ≥ Np 1 = Na 1 . As N → ∞ we see the partial sums are unbounded since a 1 > 0 and thus the original series must diverge as claimed. M1C: Sequences and Series Practice c Jeffrey A. Anderson Page 12 of 12

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