Math 22 - Review for Exam 3. Answer each of the following questions as either True or False. Circle the correct answer. (a) True/False: If a n > 0 and a n 0, the series a n converges. Soln: False: Let a n = /n. Clearly a n 0 but the series a n is the harmonic series which, by the integral test, is divergent. True/False: If a n converges, then 0 a n converges. Soln: True: One series is just a constant plus the other one. If one converges, so does the other. (c) True/False: If a n > 0 for n =, 3, 5, and a n < 0 for n = 2, 4, 6, 8,, then a n converges. Soln: False: a n = ( ) n+ satisfies the above alternating property but the series a n is equal to + + which does not converge since the terms do not go to zero. (d) True/False: The series ( sin x 2 )n converges for all x. Soln: True: The above series is a geometric series with r = sin x. Since sin x for any x, then /2 < r < /2 and so the geometric 2 series converges (for every value of x). (e) True/False: If a n converges, then a 2 n converges. Soln: False: If a n = ( )n, then by the alternating series test, the series n the harmonic series which diverges. (f) True/False: If lim b n+ n b n = 2, then b n ( 4 )n converges. a n converges. However, a 2 n = /n and so the series a 2 n is Soln: True: Consider the power series b nx n. Using the ration test to compute the radius of convergence of the power series gives us lim n b n+x n+ b nx n = x lim n b n+ = 2 x. b n Thus, by the ratio test, the radius of convergence of this power series is /2, that is to say the series converges for /2 < x < /2. Since x = /4 lies in this interval of convergence, the above series converges. (g) If the numbers a 0, a, a 2, a 3..., decrease to 0, then Soln: True: Alternating series theorem. (h) If C n ( 2 )n converges, then ( ) n a n converges. C n ( 4 )n converges. Soln: True: The hypothesis of the statement says that the power series C nx n converges when x = /2. But by a theorem about power series, the series converges for /2 < x < /2. In particular, the series converges at x = /4. (i) The series (.23) n = +.23. Soln: False: Although the series is a geometric series with r =.23 and the formula is applied correctly, the series does not converge since r >. Recall that for a geometric series to converge, the ratio r must satisfy < r <. (k) If the terms of a series alternate, then the series converges. Soln: False: The series + + alternates but does not converge since the terms do not go to zero. (l) If a n > 0 and a n < e n, then a n converges.
2 Soln: True: The series e n is a convergent geometric series and by hypothesis, a n e n. (m) /( x) = + x + x 2 + x 3 + x 4 for every x. Soln: False: This is a geometric series which converges only when < x <. (n) If a n > 0 and lim n a n+/a n =, then the series diverges. Soln: False: Consider the example a n = /n 2. It is easy to check that the hypothesis are satisfied. However, by the integral test, the series /n 2 converges. (o) If a n > 0 for all even n and a n < 0 for all odd n, then a n converges. Soln: False: If a n = ( ) n, then a n = + + which diverges. (p) If 0 < a n < and a n converges, then a 2 n converges. Soln: True: Since 0 < a n <, then a 2 n < an. Now use the comparison test. (q) The series ln 2 + ln 3 ln 4 + converges. Soln: True: Alternating series theorem. (r) The series Soln: True: Ratio test. n 2 3 n converges. (s) + e + e 2 + e 3 + = e. Soln: False: Can t use geometric series formula here since e >. (t) If a n x n converges when x = 6, then na n 4 n converges. Soln: True: The hypothesis say that the series converges in the interval ( 6, 6). The derivative of the power series will also converge in the same interval. In particular, the derivative of the power series will converge at x = 4. (u) The series n n + 2 converges. Soln: False: Compare the terms of the series with / n. The series / n diverges. (v) + + + = 0. Soln: False: The series does not converge (partial sums are either or 0) and so it can t equal anything. (w) If p is any polynomial, then the Taylor series about any point is equal to p. Soln: True 2. If a n is a sequence of numbers with 0 < a n <, which of the following conditions will guarantee that the series a n converges. Circle as many as necessary. No explanations are required. (a) a n 0 as n. a n /n 2 for all n. (c) The series (d) a n > /n for all n a n converges.
3 (e) The series a 2 n converges. Soln: (a) No. Let a n = /n. Then 0 < a n < but the series a n = ; Yes. Use the comparison and integral tests; (c) No. If the series /a n converges, then /a n 0 and so a n, which contradicts the hypothesis 0 < a n < ; (d) No. By the comparison test and integral test, this says just the opposite; (e) No. If a n = /n, then a 2 n converges. But a n does not. 3. (a) If a n 0, does this mean the series example. a n converges? If it does, show why. If it does not, give an Soln: No. The series /n has terms which do to zero. Yet, by the integral test, does not converge. What does the following series equal 2 + 3 4 +? Hint: Look at ln( + x). 5 Soln: The function /( + x) = x + x 2 x 3 + x 4 for < x <. Taking anti-derivatives of both sides of this equation give us Evaluating at x = gives us ln 2 = 2 + 3 4 + 5. ln( + x) = x x2 2 + x 3 3 x 4 4 +. (c) Does the series + + + converge? Give a reason to support your answer. Soln: No. The terms of the series do not go to zero. (d) What does the series e + e 2 + e 3 + e 4 + converge to? Soln: This series is a geometric series with r = e (the ratio) and a = e (the first term). Note that < r < and so the series converges to a/( r) = e /( e ). (e) Does the series your answer. 2 + 2 2 2 + 3 2 3 + 4 2 4 + = converge? Give a reason to support n2n Soln: Yes it does. Let a n = /n2 n and notice that lim n a n+ = <. By the ratio test, the series will converge. a n 2 4. /3 + /9 /27 + =? Soln: The above us a geometric series with (ration) r = /3 and (first term) a =. The sum is thus a/( r = /( + /3) = 3/4. 5. Discuss the convergence (or divergence) of the following series. (a) (c) n log n ( ) n n log n ( 2) n n=2 n=2 Soln: (a) Diverges (integral test) Converges (alternating series test) (c) Diverges (terms don t go to zero) 6. Discuss the convergence of divergence of the following series. (a) (.999) n ne n2
4 (c) n 0 e n Soln: (a) Converges (geometric series) Converges (integral test - or ratio test) (c) Converges (ratio test) 7. Discuss the convergence (or divergence) of the following series. (a) 2 2 + 2 2 + 2 2 + 2 2 2n (n + )! (c) 2 3n (d) n log n Soln: (a) Diverges (terms don t go to zero) Converges (ratio test - with a little algebra) (c) Converges (note that 2 3n = (/8) n and so this series is a geometric series) (c) Diverges (integral test) 8. Discuss the convergence (or divergence) of the following series. (a) 2 n (c) n ln n n 00 e n n=2 Soln: (a) Converges (geometric series) Diverges (integral test) (c) Converges (ratio test) 9. Discuss the convergence of the series ( + /n) n 2 n. Soln: Note that ( + /n) e ( + /n)n e as n. Thus 2 n C. This last quantity is the term of a geometric series. So, by comparison, 2n the series converges. 0. Does the following series converge? 2n 2 + n 3 n 3 + 2n Soln: Note that 2n2 + n 3 n 3 + 2n C. But the series n. Does the following series converge? Does it converge absolutely? ( ) n n 2 /n diverges (by the integral test) and so, by comparison, the original series diverges. 3 n Soln: The series converges since it is an alternating series and the absolute value of the terms monotonically decreases to zero as n (alternating series theorem). Taking absolute values of the terms and using the ratio test, one can show that the sum of the absolute values of the n 2 terms of the series, i.e., converges. Thus the series converges absolutely. 3n 2. Does the following series converge? Does it converge absolutely? ( ) n n log n
5 Soln: The series converges, alternating series test, but does not converge absolutely since the series n log n integral test. So the series converges but not absolutely. does not converge, by the 3. Does the following series converge? Does it converge absolutely? ( ) n n Soln: The series converges, alternating series test, but does not converge absolutely since the series does not converge, by the integral n test. So the series converges but not absolutely. 4. Discuss the convergence (tell if the series converges absolutely, conditionally, or not at all) of the following series: (a) n + 5 n cos(/n 2 ) (c) ( ) n n 2 n 3 + 2 Soln: (a) Absolutely (ratio test) Diverges (terms don t go to zero) (c) Conditionally (the series converges - alternating series test - but the absolute value of the terms is approximately C/n which form the terms of the divergent series - integral test) 5. Compute exactly the following infinite sums. Circle your answer. () 2 + 3 4 + 5... (2) 2! + 4! 6! + 8!... 2 n (3) n! (4) e n = e n Soln: () This series converges to ln 2. The power series for ln( + x) = x x 2 /2 + x 3 /3. Now evaluate both sides at x =. (2) This series converges to cos. See this by writing out the power series expansion of cos x and evaluating both sides at x =. (3) This series converges to e 2. See this by writing out the power series expansion of e x and evaluating both sides at x = 2. (3) This series converges to /( /e) (geometric series). 6. A ball is dropped from ft. off the ground. Each time it hits the ground, it bounces /3 of the height it fell. (e.g. On on the first bounce, it rises /3 ft.) If the bouncing if allowed to continue forever, find the total distance travelled by the ball. Soln: The distance travelled is + 3 + 3 + 9 + 9 + 27 + 27 + = + 2( 3 + 9 + 27 + ) = + 2 /3 /3. 7. The bob on a pendulum swings through an arc 24 cm long on its first swing. If each successive swing is 5/6 of the length of the preceding swing, find the total distance the bob travels. Soln: The total distance travelled is 24 + 5 6 24 + 5 5 6 6 24 + 5 5 5 6 6 6 24 + = 24( + 5 6 + ( 5 6 )2 + ( 5 6 )3 + ) = 24 5/6. 8. What is a an e n!? Soln: Notice that e a a n = and so n! a an e = e a a n = e a e a =. n! n! 9. Use the power series for sin x to write I = 0 sin xdx as an infinite series. Now use the theory of alternating series to determine how many terms of this series are needed to estimate I within 0 3. Then compute this series estimate of I.
6 Soln: sin x = x x3 3! + x 5 5! x 7 7! +. Thus sin x dx = 0 2 4! + 6! 8! +.. This is an alternating series and so the error is at most the absolute value of the (n + )st term. One checks that /8! < 0 3 and so sin x dx 0 2 4! + 6! with an error no more than 0 3. 20. (a) Approximate.2 by using the third degree Taylor polynomial (about x = 0) of + x. Soln: The third degree Taylor polynomial for this function is + x + x 2 x2 8 + x 3. Evaluate this polynomial at x = 0.2 to get 6.2.0955. Estimate the error. Soln: The error is at most max 4! f (4) (x), where f(x) = + x. Note that f (x) = 5 0 x 0.2 6 ( + x) 7/2 and so max 4! f (4) (x) 0 x 0.2 4! 5 6 = 0.0390625. 2. (a) Compute the power series for + x 2 Use this to compute the power series for tan x. (c) What is 3 + 5 7 + 9 +? Soln: The function arctan x is the anti-derivative of /( + x 2 ). The function /( + x 2 ) can be written as + x 2 = (x2 ) + (x 2 ) 2 (x 2 ) 3 + (x 2 ) 4 = x 2 + x 4 x 6 + x 8. Taking anti-derivatives of both sides gives us arctan x = x x3 3 + x 5 5 x 7 7 +. Evaluating at x = gives us 3 + 5 7 + 9 + = arctan = π/4. 22. For what x does the following power series converge (specify where is converges and where it converges absolutely). 4 n x n n. Soln: Let a n = 4n x n n. Then a n+ a n n = 4 x. Taking limits to both sides of the above gives us n + lim a n+ n = 4 x lim n a n n n + = 4 x. This limit is less than one when x < /4. Thus the series converges (absolutely) for /4 < x < /4. When x = /4, the series becomes ( ) n which diverges (integral test). When x = /4, the series becomes which converges (alternating series test) but not absolutely n n (integral test). 23. For what x does the following power series converge (specify where is converges and where it converges absolutely). x n 2 n n. Soln: Let a n = xn and note that lim n2n n a n+ x n = lim a n n 2 n + = x. Thus, by the ratio test, the series converges absolutely for x < 2. 2 What about the endpoints? When x = 2, the power series becomes which diverges (integral test) while when x = 2, the series becomes n ( ) n which converges (alternating series test) but not absolutely (integral test). n 24. Compute the radius of convergence of the power series Soln: Let a n = 2n (n!) 2 xn and use the ratio test. 2 n (n!) 2 xn. a n+ = 2 n+ x n+ a n ((n + )!) 2 (n!) 2 2 n x n = 2 (n + ) 2 x. Taking limits as n we get this ratio is zero. Thus, by the ratio test, the radius of convergence is infinity. That is to say, the series converges for all values of x.
7 25. Find the radius of convergence of the following power series. 2 n (a) (n!) 2 xn Soln: If a n = 2n a n+ (n!) 2 xn, then a computation reveals that lim = lim n a n n x 2 = 0 for all x. Thus, by the ratio test, the series (n + ) 2 converges for all x. (n + )!x n Soln: If a n = (n + )!x n, a computation reveals that only for x = 0. a n+ lim = lim (n + 2) x = for all x. By the ratio test, the series converges n a n n 26. The two power series a n (x 2) n and b n (x 3) n both converge when x = 6. Find the largest interval for which both series must converge. Soln: The first series will converge in the interval ( 2, 6]. The second will converge in the interval (0, 6]. The intersection of these two intervals is (0, 6]. 27. Suppose f(x) = a n x n has a radius of convergence r; i.e., the power series converges to f(x) for r < x < r and does not for x > r. What is the radius of convergence for the series f(bx) = a n (bx) n? Soln: If the above series has radius of convergence r, then the series f(x) converges for x < r and diverges for x > r. It follows that the series f(bx) converges for bx < r, i.e., x < r/b, and diverges for bx > r, i.e., x > r/b. Thus the radius of convergence of f(bx) is r/b. What is the radius of convergence for the series f(x 2 ) = a n x 2n? Soln: By the reasoning above, the series f(x 2 ) converges for x 2 < r, i.e., x < r and diverges for x 2 > r, i.e., x > r. Thus the radius of convergence of f(x 2 ) is r. 28. Suppose that the power series a k (x 2) k converges when x = 5 and diverges when x = 7. Answer k=0 the following questions with short answers. (a) Does the series converge when x = 0? Does the series ka k (x 2) k converge when x = 0? k= (c) Does the series converge when x = 20? (d) Is the radius of convergence of the series equal to 3? (e) Does the series converge when x =? Soln: First a general observation. From the theory of power series the interval of convergence must be symmetric about the base point x = 2. Thus the series converges for < x 5 and diverges for x 7 and for x >. From here we can get the answers to the questions. (a) Yes; Yes (this series is the derivative of the given series which will have the same interval of convergence); (c) No; (d) The radius of convergence is at least 3, but it may be more. Not enough information to tell; (e) Not enough information to tell. 29. Recall that the interval of convergence is the set of all points for which a power series converges.
8 (a) Find the interval of convergence for ( ) n 3 n x n n 4. Soln: Using the ratio test, we see that the series converges when 3 x <, i.e., when /3 < x < /3. At the endpoint x = /3, the series ( ) n is n 4 which is a convergent alternating series. At the other endpoint x = /3, the series is n 4 which is convergent by the integral test. Thus the interval of convergence is [ /3, /3]. Find a power series whose interval of convergence is (, ]. ( ) n x n Soln:. n 30. Determine all x for which the series 4 n (x 3) n n converges. Soln: If a n = 4n (x 3) n a n+ n, then a computation reveals that lim = lim n n a n n 4 x 3 = 4 x 3. This is less than one when n + x 3 < /4 which holds when 2.75 < x < 3.25. So by the ratio test, the series converges when 2.75 < x < 3.25 and diverges when x > 3.25 ( ) n or when x < 2.75. When x = 3.25, the series becomes which is convergent (alternating series theorem). When x = 3.25, the series n becomes which diverges (integral test). n 3. (a) Write down the first 5 terms in the power series expansion of Soln: First notice that = + u u 2 + 3 8 u2 + 5 6 u3 + 35 x 2 28 u4 +. Now let u = x 2 in the above power series to get = + x 2 2 x2 + 3 8 x4 + 5 6 x6 + 35 28 x8 +. Write down the first 5 terms in the power series expansion of sin x. Soln: Integrate both sides of what we just computed to get sin x = x + 6 x3 + 3 40 x5 + 5 2 x7 + 35 52 x9 + 32. Compute the first 8 terms in the power series expansion of f(x) = tan (x) about x = 0. Soln: First note that tan x is the anti-derivative of Now integrate to get tan x = x x3 3 + x 5 5 x 7 7 +. + x 2. Next note that + u = u + u2 u 3 + u 4 and so + x 2 = x2 + x 4 x 6 + x 8. 33. Compute the first 8 terms in the power series expansion of f(x) = ( + 2x) 2 about x = 0. Soln: Notice that ( + u) 2 = d du + u. Also notice that + u = u + u2 u 3 + and so Finally, setting u = 2x we get ( + 2x) 2 = 4x + 2x2 32x 3 +. ( + u) 2 = 2u + 3u2 4u 3 + 5u 4. 34. Compute the Taylor polynomial P 3 (x, 0) for the function f(x) = log( + x). Soln: First note that + x = x + x2 x 3 + x 4 + and so, taking anti-derivatives of both sides, we get log( + x) = x x2 2 + x 3 3 x 4 4 +. 35. Suppose f(x) has power series x + 4 x2 + 9 x3 + = n 2 xn.
9 (a) What is f (20) (0)? Soln: By the theory of power series, the constant in front of x 20 is f (20) (0). In the power series above, the constant in front of x 20 is 20! 20 2. Thus f (20) (0) 20! = 20 2 for which follows that f (20) (0) = 20! 20 2. Is f(x) defined at x =? Explain. Soln: Yes, since, by the integral test, the series n 2 converges. (c) Is f(x) defined at x = 2? Explain. 2 n Soln: No, since, by the ratio test, the series n 2 does not converge. 36. What degree Taylor polynomial about x = 0 is necessary to approximate cos x in the interval [ 2, 2] with an error less than 0 6. Soln: The standard error estimate says that E N (x) max f (N+) (t) x N+ N+ 0 t x (N + )! 2 (N + )!. Note that 2 4 /4! =.879 0 7 and this is the smallest N that works. Thus we need to take the 3-th degree Taylor polynomial of cos x. 37. (a) Compute the 4th degree Taylor Polynomial of e x. Soln: e x + x + x2 2! + x 3 3! + x 4 4!. Estimate e.0. Soln: e 0.0 + ( 0.0) + ( 0.0)2 (c) Estimate the error. 2 + ( 0.0)3 6 + ( 0.0)4 24 0.99005. Soln: If f(x) = e x, the error E 4 satisfies the inequality E 4 max 0.0 x 0 5! f (5) (x) 0 ( 0.0) 5 = 5! (0.0)5 = 8.333 0 3. (d) How many terms are needed for the error to be less than 0 7? Soln: In general, the error is estimating e 0.0 using an n-th degree Taylor polynomial is (using the a similar calculation as above, is E n (0.0)n+. One quickly checks, by trial and error, that when n = 3 that (0.0) n+ /(n + )! < 0 7. (n + )! 38. Estimate e /4 using P 2 (x, 0) and estimate the error R 2 (x, 0). Soln: This problem is similar to the one above. 39. Estimate sin(/2) with P 5 (x, 0) and estimate the error. Soln: sin x x x3 6 + x 5 20 and so sin 2 2 ( 2 )3 6 + ( 2 )5 inequality E 5 max 6! f (6) (x) 0 /2 6 0 x /2 6! ( 2 )6 = 0.00002704 20 = 0.479427. Let f(x) = sin x and note that the error E 5 satisfies the 40. Use the fourth degree Taylor polynomial (i.e. P 4 ) to approximate.2 and estimate the error. Soln: Note that x + x 2 + 3 8 x2 5 6 x3 + 35 28 x4. Now evaluate this at x = 0.2 to get.2 0.99048. If f(x) = / + x, note that the error E 4 is satisfies the inequality E 4 max 5! f (5) (0) 0 0.2 5 = 945 max 0 x 0.2 5! 0 x 0.2 32 ( + x) /2 (0.2)5 = 945 5! 32 (0.2)5 = 0.00007875 4. Approximate the following integral using power series and explain what you are doing /2 0 x 3 x 3 dx.
0 Thus Soln: Note that u = u 2 u2 8 u 3 6 5 28 u4 and so x 3 = x 3 2 x 6 8 x 9. From this we get that 6 x 3 x 3 = x 3 x6 2 x 9 8 x 2 6. /2 x 3 x 3 dx = (/2)4 0 4 (/2)7 4. 42. How many terms of the power series for e x (about x = 0) are needed to approximate e 0. to within 0 3? Soln: This problem is similar to one presented earlier. 43. Compute the following limits: cos x (a) lim x 0 + x cos x lim x π/2 (x π/2) 44. Compute the following limits e 2x 2 x 0 x 2 (a) lim lim x 0 e x x x 2 cos 2x 45. Using Taylor series, compute the following limits: 4x sin(4x) (a) lim x 0 x 3 lim x 0 (sin x x) 3 x( cos x) 4 46. If f(x) = e 2x2 x 2, compute f (6) (0). 47. (a) Evaluate the following limit e 2x lim x 0 tan x. Soln: First note that e 2x = 2x + 2x 2 4 3 x3 + and tan x = x x3 3 + x 5. Thus 5 Taking limits as x 0 gives us 2. e 2x 2 2x + 2x 4 tan = 3 x 3 + x( 2 + 2x 4 3 = x2 + ) 2 + 2x 4 3 = x2 +. x x x3 3 + x5 5 x( x2 3 + x4 5 ) x2 3 + x4 5 If f(x) = tan x, what is f (5) (0)? Soln: Note that tan x = x x3 3 + x 5 5. Also note that the coefficient in front of x5 (namely /5) is equal to f (5) (0)/5!. Thus f (5) (0) = 5!/5 = 4! = 24. 48. If f(x) = cos(x 2 ), compute f (00) (0). 49. (a) In planning a highway across a desert, a surveyor must make compensations for the curvature of the earth when measuring differences in elevation. Using trigonometry one proves that if s is the length of the highway and R is the radius of the earth, then the correction C is given by C = R(sec(s/R) ).
If f(x) = sec x, the use the following table to get an estimate of C in terms of s and R. Compute the following limit f(0) f (0) f (0) f (0) f (0) 0 0 5 sec( 2x 2 ) lim x 0 x 4. 50. Suppose we are given a function f(x) such that f(0) =, f (0) = 2, f (0) = 3, f (0) = 4, and f (iv) f(x) (0) = 5. Compute lim. x 0 x 5. Find the sum of the following series (a) ( ) n x4n n! ( ) n π 2n+ 4 2n+ (2n + )! ( ) n π 2n (c) 6 2n (2n)! 52. Evaluate exactly the following series. (a) n!2 n. Soln: This series is equal to x n /n! = e x when x = /2. Thus the series sums to e /2. n 3 n. Soln: This series is equal to /3 nx n = d 3 dx x = 3 when x = /3. This evaluates to 3/4. ( x) 2