Calculus II Review Solutions 2003

Does it converge or diverge? If it converges, find its value (if possible). 1. ∞ X 1 √ n− n n=2 P1 The terms of the sum go to zero. It looks similar to n , which diverges. We also note that the terms of the sum are positive. We compare them: 1√ n− n lim 1 n→∞ n 2. n 1 √ = lim =1 n − n n→∞ 1 − √1n P The series diverges by the limit comparison test, with (1/n). n o = lim n→∞ n√ 1+ n In this case, we simply take the limit: n √ = lim n→∞ 1 + n n→∞ lim √ √1 n n =∞ +1 The sequence diverges. 3. ∞ X n2 + 1 n3 − 1 n=2 2 The terms of the sum go to zero, since there is an Pn1 in the numerator, 3 and n in the denominator. In fact, it looks like n , so we compare it to that: n2 −1 n3 − n 3 =1 lim n 1−1 = lim 3 n→∞ n − 1 n→∞ n P1 Therefore, the series diverges by the limit comparison test, with n. √ ∞ X 5−2 n 4. n3 n=1 We can temporarily break this apart to see if the pieces converge: √ ∞ ∞ ∞ √ X X 5−2 n X 5 n = −2 3 3 3 n n n n=1 n=1 n=1 Both of these are p−series, the first with p = 3, the second with p = 52 , therefore they converge separately, and so the sum also converges. 5. ∞ X (−6)n−1 51−n n=1 First, let’s rewrite the terms of the sum: n−1 1−n (−6) 5 (−6)n−1 = = 5n−1 1  −6 5 n−1 6. so that this is a geometric series with r = diverges. n o −6 5 . Since |r| > 1, this series n! (n+2)! We first simplify: n! 1 = (n + 2)! (n + 1)(n + 2) so the limit as n → ∞ is 0.   ∞ X n 7. ln n+1 n=1 A little tricky... First, note that this can be written as ln(n) − ln(n+1 ). Now, let’s write out the nth partial sum: Sn = ln(1) − ln(2) + ln(2) − ln(3) + ln(3) − ln(4) + . . . + ln(n) − ln(n + 1) with cancellations, Sn = 0 − ln(n + 1) Now, the limit of Sn as n → ∞ is −∞, so the sum diverges. 8. ∞ X 3n + 2n 6n n=2 A sum of geometric series: ∞ ∞  n ∞  n X X X 3n + 2n 1 1 (1/2)2 (1/3)2 2 = + = + = n 6 2 3 1 − (1/2) 1 − (1/3) 3 n=2 n=2 n=2 9.  sin nπ 2
Write out the first few terms of the sequence: 1, 0, −1, 0, 1, 0, −1, . . . so the sequence diverges. 10. ∞ X 1 n(n + 1)(n + 2) n=1 First, we see the terms go to zero like 1 n3 . n3 =1 n→∞ n(n + 1)(n + 2) lim so the series converges by the limit comparison test. 2 11. ∞ X sin2 (n) √ n n n=1 First, do the terms go to zero? The maximum value of the sine function is 1, and all terms of the sum are positive, so: sin2 (n) 1 ≤ 3/2 3/2 n n so the terms do goP to zero. Actually, we’ve also done a direct comparison ∞ 1 with the p−series n=1 n3/2 , which converges. 12. ∞ X n (n + 1)2n n=1 It looks like the terms are going to zero like 21n , so let’s compare it to P (1/2)n , which is a convergent geometric series. n 1 1 · n ≤ n n+1 2 2 So the series converges by a direct comparison. Evaluate, if possible. Z x3 1. dx 3 x +1 Do long division first! 1 x3 =1− 3 3 x +1 x +1 Can we factor x3 + 1? We see x = −1 gives 0, so x + 1 can be factored out. Using long division, x3 + 1 = x2 − x + 1 x+1 so that x3 + 1 = (x + 1)(x2 − x + 1) (NOTE: On the exam, you will be able to factor the polynomial easier than this!) By Partial Fractions, x3 1 1 1 1 2−x =1− =1+ · + · 2 3 2 x +1 (x + 1)(x − x + 1) 3 x+1 3 x −x+1 Now you have to complete the square to finish things off, and after some long algebra,   (2x − 1) 1 1 1 2 −1 √ − ln(x + 1) x + ln(x − x + 1) − √ tan 6 3 3 3 NOTE: This was a complicated exercise! If you made it through this one, you could probably stop now- You’re ready! There won’t be anything this complex on the exam... 3 Z 2. 0 1 1 dx 2 − 3x There is a vertical asymptote at x = 23 , so we need to split the integral there: Z T Z 1 1 1 1 dx = lim − dx + lim + dx T →2/3 T →2/3 2 − 3x 0 2 − 3x 0 2 − 3x Integrate by taking u = 2 − 3x, and we get that the antiderivative is − 31 ln |2 − 3x|. Now, take the limits- we’ll do one here:   1 1 1 1 1 1 1 lim − − ln + · = lim − + ln (2 − 3T ) + · 3 2 − 3T 3 2 T →2/3 3 3 2 T →2/3 which diverges, since 0 is a vertical asymptote for ln(x). Z 1 3. dx x4 − x2 Factor and use partial fractions: Z 1 1 1 1 dx = + ln(x − 1) − ln(x + 1) x2 (x − 1) x 2 2 Z 4. ∞ 1 dx 1 + ex 1 Use u = 1 + ex , so du = ex dx, so that du = (u − 1)dx. Substitution gives: Z 1 du = u(u − 1) Z −1 1 + du u u−1 Antidifferentiate, and we get: − ln (1 + ex ) + ln(ex ) = ln  ex 1 + ex  Take the appropriate limit to get an answer of ln(2) Z ∞ dx 5. dx 2 (x + 2) (x + 1) 0 Use partial fractions: Z Z dx 1 1 1 dx = + − dx 2 2 (x + 1) (x + 2) (x + 1) x+2 x−1 Antidifferentiate to get: − 1 1 x+2 + ln |x + 2| − ln |x + 1| = − + ln x+1 x+1 x+1 And, take the limit to get −1 + ln(2) 4 5x2 + 3x − 2 dx x3 + 2x2 Use partial fractions to get: Z Z 6. −1 2 3 + + dx 2 x x x+2 And integrate to get: 1 + 2 ln(x) + 3 ln(x + 2) x 5