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EXAM 3, SOLUTIONS Math 1220-003 - Calculus II

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EXAM 3, SOLUTIONS Math 1220-003 - Calculus II EXAM 3, SOLUTIONS Math 1220-003 - Calculus II Problems 1. Find the limit or say it does not exist, et − 1 lim t!0 t3 Solution. This is an indeterminate form of type 0/0 (when plugging in 0 for t directly). Thus L'Hopital's rule must be applied: et − 1 et lim = lim t!0 t3 t!0 3t2 At this point, notice that plugging in 0 gives 1/0, which is no longer an indeterminate form. In fact, since 3t2 is always positive, the second limit is approaching 1 as t approaches 0. Thus by L'Hopital's rule, the original limit is also 1, or DNE. Note that this example illustrates it is incorrect to continue applying L'Hopital on a form that is not indeterminate. Many students kept applying L'Hopital to arrive at et 6 , stating as the answer 1/6. You MUST check after each application of L'H that it is still indeterminate. 2. Find the limit or say it does not exist, lim x1=x x!1 Solution. This is indeterminate of type 10. Any exponential indeterminate type calls for use of logarithms: L = lim x1=x x!1 ln L = lim ln x1=x x!1 ln x ln L = lim x!1 x 1 The limit on the right is now an indeterminate of type 1 so L'Hopital may be applied: 1 1=x ln L = lim x!1 1 1 ln L = lim x!1 x ln L = 0 L = e0 = 1 3. Compute the following integral or say that it diverges, Z 9 dx 1=3 1 (x − 1) Solution. The integral is improper because the integrand increases to 1 near one of the limits of integration (x = 1). Thus it should be rewritten, Z 9 dx lim 1=3 a!1 a (x − 1) Now compute the integral as normal, first substituting u = x − 1; du = dx: Ignore the limits of integration for now, until after plugging x's back in later. du lim a!1 (u)1=3 3 lim[ u2=3] a!1 2 3 2=3 9 lim[ (x − 1) ]a a!1 2 3 lim [(9 − 1)2=3 − (a − 1)2=3] a!1 2 3 (4) = 6 2 2 4. Find the sum of the following infinite series, or say it diverges: 1 X 2n 3n+1 n=1 Solution. First rewrite the given series as 1 1 X 2n 3 3n n=1 1 1 X 2 ( )n 3 3 n=1 It is now clearly a geometric series with first term a = 2=9 and ratio r = 2=3. Thus, a 2=9 2 3 2 S = = = · = 1 − r 1 − 2=3 9 1 3 5. Find the sum of the following infinite series, or say it diverges: 1 X n2 n2 + 1 n=1 The limit of the terms at 1 is n2 lim = 1 n!1 n2 + 1 by comparing coefficients on the highest powers of n in the top and bottom. Thus by the nth Term Divergence Test, since the limit of the terms is not 0, the series diverges. 6. Use the Integral test to show whether the following series converges or diverges: 1 X 1 n2 n=1 Note: Since I asked you to use the Integral Test, I marked off points for only stating the p-series result without any other work. To use the Integral Test, set up the integral Z 1 1 2 dx 1 x 3 Z t 1 lim 2 dx t!1 1 x 1 t lim [− ]1 t!1 x 1 1 lim [− + ] = 1 t!1 t 1 Since the integral converged, the series converges also. Note. The Integral Test does NOT say that the series converges to 1. It only tells whether it converges or not (I mentioned in class that this series in fact converges to π2=6) 7. Determine whether the given series converges or diverges. You may use the result of the previous problem without having to prove it again. 1 X n + 1 n3 − n n=1 Note. "The Hierarchy" is not a valid answer to a problem like this! I have included a note at the end of the test in regards to this. Note. This problem had an unintended error on my part, and that is that the first term in the series is in fact 2/0, which does not exist (so the series technically diverges!) I should have given full points to anyone that came to this conclusion by either check- ing the first term or using the integral test (which would have given divergence). Solution. What the Hierarchy does help you do is determine what function to use in the comparison test. Ignoring all but the highest power of n in top and bottom, our P P n P 1 suggested comparison series bn is n3 , or n2 . We proved in the previous problem that this series converges, so if the Limit Comparison Test gives a finite number, the series we're interested in converges also. a lim n n!1 bn (n + 1)=(n3 − n) lim n!1 1=n2 4 n + 1 n2 lim · n!1 n3 − n 1 n3 + n lim = 1 n!1 n3 − n Thus by the limit comparison test the original series converges (minus the note above). Note that the problem may have been simplified slightly from the beginning by factor- P 1 ing the bottom and cancelling n + 1 from the top and bottom, leaving n2−n 8. Determine whether the given series converges or diverges. Explain your answer. 1 X n − 1 4n n=1 Note. Same as previous problem, "The Hierarchy" does not suffice as an answer here. See note at the end of the test. Solution. The exponential term suggests using the Ratio Test. a lim n+1 n!1 an (n − 1 + 1)=(4n+1) lim n!1 (n − 1)=(4n) n 4n lim · n!1 4n+1 n − 1 n 1 lim = n!1 4(n − 1) 4 1 Since 4 < 1, the ratio test says that the series converges. Note. The test does NOT say that the series converges to 1/4. n Note. Some students used the Limit Comparison Test to compare the series to 4n and got a ratio of an = 1. This does NOT say that your original series converges! It only bn says that your original series does the same thing as the new one. So if you use this P n test you still have to prove that 4n converges also, which will require the Ratio Test similar to above. 5 9. Determine whether the given series converges or diverges. If it converges, specify whether it converges conditionally or absolutely. 1 X 2n + 1 (−1)n+1 3n − 1 n=1 The series diverges by the nth Term Divergence Test: 2n + 1 lim = 2=3 n!1 3n − 1 The (−1)n+1 will simply throw in negative signs every other term and in fact the limit of the terms DNE (since they bounce back and foth between 2/3 and -2/3 in the limit). Bonus Problems P P P P 1. Give an example of two series an and bn such that an and bn converge, but P anbn diverges. Solution. As Problem 2 below suggests, one or both of the series we find here should have negative terms in it. The easiest example is something of this sort: 1 a = b = (−1)n+1 p n n n P P The series an = bn converge by the Alternating Series Test since the terms go to 0, alternate, and are decreasing. However, X X 1 1 a b = (−1)n+1 p (−1)n+1 p n n n n X 1 = (−1)2n+2 p p n n X 1 = 1 · n This is now the harmonic series (with all positive terms), which diverges. P P 2. Show that for ANY series an and bn which are POSITIVE and converge (an; bn > P 0 for all n) then anbn must converge also. Solution. The easiest way to get to the finish is the following: 6 P Since an converges, its terms must go to 0, that is, lim an = 0 n!1 By the definition of the above limit, there is an index N such that an < 1 for all n > N. Now by the Ordinary Comparison Test, 1 1 X X anbn ≤ 1 · bn N N P The right sum is part of the series bn, which we assumed converges. Thus the left sum must also converge. Furthermore, adding in the first N − 1 terms to the left sum will not affect convergence of the series since this is a finite number of terms. Thus P anbn must converge, which is the desired result. Question. Where does this argument break down if the series an and bn are not both positive? Additional Notes on the Test THE HIERARCHY In class I gave a hierarchy of how functions act against each other at 1: nn > n! > 1000n > 2n > n1000 > n2 > n1=1000 > ln(n) > 1 What the Hierarchy means is the following: 1) If you are finding a limit at infinity of an , the Hierarchy tells you what the result bn will be without any further work. If an > bn on the hierarchy, the result will be 1. If an < bn on the hierarchy, the result will be 0. 2) If you are determining whether a series converges/diverges, you may use the hi- erarchy along with the Limit Comparison Test.
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