Math WW08 Solutions November 19, 2008

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1 Math 352- WW08 Solutions November 9, 2008 Assigned problems 8.3 ww ; 8.4 ww 2; 8.5 4, 6, 26, 44; 8.6 ww 7, ww 8, 34, ww 0, 50 Always read through the solution sets even if your answer was correct. Note that lie many of the integrals in this course, there is frequently more than one way to determine convergence or divergence of a series. Your solution may be correct even if you used a different method than what I use here.. (Note, webwor creates slight variations on this problem for each student. So your problem may be slightly different.) N 3x 2 e x3 dx 3x 2 e x3 dx N We ll do a simple u-substitution. Let u x 3, du 3x 2 dx N xn x e u du xn N e u x e x3 N N 3 N e N + e ( N e ) e N e e So the improper integral converges to /e. Next, we are ased to determine whether the following series converges. 3n 2 e n3 n Given the integral we just did, it s a rather big hint that we might want to try the integral test. In order to apply the integral test, let f(x) 3x 2 e x3. The integral test applies if f(x) is positive, continuous, and decreasing. So first we need to chec that we ve met the conditions of the test. The function f(x) is clearly positive since x 2 is positive and e to any power is also positive. It is continuous since it s the product of continuous functions (nown from Calc. I). So the only thing left to show is that f(x) is decreasing. A function is decreasing if its derivative is negative. f(x) 3x 2 e x3 f (x) 6xe x3 + (3x 2 )(e x3 )( 3x 2 ) (Product rule and chain rule.) 6x 9x4 e x3

2 Math 352- WW08 Solutions November 9, 2008 The denominator is always positive. The numerator is negative for all x > 3 2/3. So f(x) is eventually decreasing. Therefore, the integral test applies, and the series and the improper integral either both converge or both diverge. The integral converges, so the series converges. 2. (Note that webwor creates random, but similar, problems for each student on this question. So the solutions here are for a sample of problems to give you some examples. If you have questions on any particular problem you were given, please as. (2.) n 7n6 n 8 +3 n a n Since the series loos almost lie a p-series (except for the +3 in the denominator it loos lie a p-series with p 2), I ll try it comparison test. The given a n terms are similar to b n 7n6 n 8. a n 7n 6 /(n 8 + 3) n b n n 7n 6 /n 8 7n 6 n 8 n 7n 6 (n 8 + 3) n + 3/n 8 The it is positive and finite (i.e., 0 < < ) therefore the a n and b n series converge or diverge together. 7n 6 n 8 7 n 2 n n is a convergent p-series (p 2 > ), therefore converges. So the original series also converges by the it comparison test. (2.2) 7n 8 n 7 +7 n n 4n 0 n 6 +3 n a n Trying the it comparison test. The given a n terms are similar to b n 7n8 4n. 0 a n (7n 8 n n)(4n 0 ) n b n n (4n 0 n 6 + 3)(7n 8 ) (28n 0 4n n 5/2 ) n (28n 0 7n 6 + 2) (28 4/n + 28/n 2/5 ) n (28 7/n 4 + 2/n 0 ) The it is positive and finite, so a n and b n converge or diverge together. bn 7 4 n 2 is a convergent p-series (p 2 > ). Therefore the original series also converges. 2

3 Math 352- WW08 Solutions November 9, 2008 (ln n) 3 n (2.3) n+3 n a n The numerator loos pretty large compared to the denominator, so I would try the divergence test first. In this case n a n 0, so the divergence test doesn t tell us anything. Next I ll try a it comparison with b n n, which is a divergent p-series (p ). a n (ln n) 3 /(n + 3) n b n n /n n(ln n) 3 n n + 3 n (ln n) 3 + 3n(ln n) 2 (/n) n (ln n)3 + 3(ln n) 2 The it goes to infinity, so the it comparison test doesn t apply. However, the zero-infinity version of the it comparison test does. The it of a n /b n and b n diverges, therefore a n also diverges. (2.4) n 7n6 n 7 +3 n a n This series loos similar to b n 7n 6 /n 7 7/n, which is a divergent p-series. So I ll compare with that. a n 7n 6 /(n 7 + 3) n b n n 7/n n 7 n n n + 3/n 7 The it is positive and finite, so by the it comparison test, a and b converge or diverge together. b is a divergent p-series, so a diverges. (2.5) cos 2 (n) (n) n n 6 0 cos2 (n) (n) n 6 n a n. Note that 0 cos 2 n. Therefore, n n 6 n /2 So 0 a b and b is a convergence p-series (p /2 > ). Therefore by the direct comparison test, a converges. 3

4 Math 352- WW08 Solutions November 9, (8.5 #4) n! 2 I almost always try the ratio test when there is a factorial. So I ll try that. a n+ (n + )!/2 n+ n a n n n!/2 n 2 n (n + )! n 2 n+ n! (Notice that (n + )! (n + )n! and 2 n+ 2 n 2 ) 2 n (n + ) n! n 2 2 n n! n n + 2 According to the ratio test, if the it is > (including ), the series diverges. So this series diverges. 4. (8.5 #6) n 3! Another factorial, so I ll try the ratio test again. a /( + )! a 3 /! 3 +! 3 ( + )! 3 3! 3 ( + )! 3 3! 3 ( + )! The it is less that, so the series converges by the ratio test. 5. (8.5 #26) n ( ) 2+ Since the terms are all to the th power, I ll try the root test. a ( 2 + ( ) / ) /2 The it is less than, so the series converges by the root test. 4

5 Math 352- WW08 Solutions November 9, (8.5 #44) n (ln ) Again, terms are to the th power, so I ll try the root test. a ln (ln ) 0 The it is 0, which is less that, so the series converges. 7. (Note that webwor creates random, but similar, problems for each student on this question. So the solutions here are for a sample of problems to give you some examples. If you have questions on any particular problem you were given, please as. (7.) ( ) n e n n n! The is an alternating series, so we ll try the alternating series test. So we need to show that the it of the terms a n goes to zero and that the terms are decreasing. We can show that a 0 using the squeeze theorem (from calc ). 0 en n! e n n(n )(n 2) (3)(2)() e n [n(n )(n 2) ( n 2 )][( n 2 )( n 2 2) (3)(2)()] (Assuming that n is even, I ve just split the factorial in half at the term n/2.) e n [n(n )(n 2) ( n 2 )][( n 2 )!] ( n 2 )! e n ( n ) n/2 2 ( n 2 )! en n 2 ln(n/2) ( n 2 )! (which goes to 0 as n ) We could mae a similar argument for odd n. Therefore, the a n terms are squeezed between two sequences that go to zero, so n a n 0. 5

6 Math 352- WW08 Solutions November 9, 2008 (Note that the part above is messier than I would give on an exam). Now we need to show that a n+ a n. e n+ (n + )! en n!? n!e n+ (n + )!e n? e (n + )? e n +? Yes, this is true for all n > 2. Therefore this alternating series converges. (7.2) ( ) n n n n n! First we ll chec if the terms of the series go to zero. n n n! (See Ex. 6 on p. 499 of the Strauss text for this it.) We now from this that the series diverges, since it fails the divergence test. However, the alternating series test only applies if this it is zero, so the webwor answer is N. (7.3) n ( ) n n 2 +5 First we ll chec if the terms of the series go to zero. n So the first condition of the alternating series test is met. Next we need to chec if a n+ a n. (n + ) 2 > n 2 (n + ) 2 + > n 2 + (n + ) 2 < + 5 n a n+ < a n Both conditions of the alternating series test are met, so the series converges. (7.4) ( ) n n! n n n First we ll chec if the terms of the series go to zero. n! n n 0 (See Ex. 6 on p. 499 of the Strauss text for this it.) Next we need to chec if a n+ a n. (n + )! (n + ) n+ (n + )n! (n + ) n (n + ) n! (n + ) n n! n n 6

7 Math 352- WW08 Solutions November 9, 2008 So a n+ a n. Both conditions of the alternating series are met, so the series converges. (7.5) ( ) n n! n n n The it of a 0 (which we now because it is the reciprocal of the it in (7.)). So the alternating series test does not apply and the webwor answer is N. Note, however, that the it not being zero means that the series diverges by the divergence test. (7.6) ( ) n cos(nπ) n n 5 Note that cos(nπ) is for odd n and for even n. Since ( ) n ) is also for odd n and for even n, the terms of the series are always positive. So this is not an alternating series, and the alternating series test does not apply. 8. (Note that webwor creates random, but similar, problems for each student on this question. So the solutions here are for a sample of problems to give you some examples. If you have questions on any particular problem you were given, please as. (8.) ( ) + n 2 + We ll chec first if the series converges absolutely, so we ll loo at the convergence of a 2 + This is very similar to the divergent p-series, so we ll do a it comparison with that series. /( 2 + ) / The it is positive and finite, so by the it comparison test the two series converge or diverge together. So the series diverges. The series is not absolutely convergent, so we still need to chec conditional convergence. Since it s an alternating series, we ll try the alternating series test. So we need to chec that the n a 0 and a + a. a (L hopital) 7

8 Math 352- WW08 Solutions November 9, 2008 Now we just need to chec that a + a. Is + ( + ) 2 + ( + )/(( + ) 2 + ) /( 2 + ) ( + )( 2 + ) ( + ) 2 + )?? 2 +? ? ? ? 2 +? This is true for all. The conditions of the alternating series test are met, so this series converges. It does not converge absolutely, so it converges conditionally. (8.2) ( ) + 2 n e I ll chec absolute convergence first using the generalized ratio test. a + a e ( + ) 2 e + 2 ( + ) 2 e 2 /e ( ) +2 ( + ) 2 /e + ( ) + ( 2 /e ( e 2 ( + 2/ + / 2 e The it is <, so by the generalized ratio test, the series converges absolutely. (8.3) ( ) + n 2+ This series loos lie it might fail the divergence test, so I will try that first. a / / The it a 0 therefore the series must diverge. 8

9 Math 352- WW08 Solutions November 9, 2008 (8.4) ( ) + n 3/2 This series is an alternating p-series with p 3 2 p > 0, so it converges at least conditionally. To determine absolute convergence, we loo at ( ) + 3/2 3/2 > 0. We proved in class that alternating p-series converge for This is a p-series with p 3/2. A p-series with p > converges, therefore this series converges. This means that the original series converges absolutely. (8.5) ( ) +! n ln The factorial always maes me thin to try the ratio test. generalize ratio test. 9. (8.6 #34) a + a ( ) + ( + )! / ln( + ) ( )! / ln ( + )! / ln( + )!/ ln ln ( + )! ln( + )! ln() ( + ) ln( + ) ( + ) + ln() + [ ( + ) 2 (L hopital) ] + ( + ) ln The it is >, therefore according to the generalized ratio test, the series diverges. So I ll test for absolute converges first with the (a.) (b.) The first for terms S 4 of the alternating series give S The error in this estimate from taing 4 terms is S S 4 a So the error is

10 Math 352- WW08 Solutions November 9, 2008 (c.) How many terms for 3-place accuracy? We need S S N a N+ < (N + ) (N + ) N So we need N 44 or at least 44 terms of the series for 3 digits of accuracy. 0. This series is the alternating series So we need S S N a N+ < N N+ 0 3 N + 3 N 2 0 ( ) 0 So for this accuracy we need N 2 or at least S 2. Note that the series starts with N 0 so S 2 is the first 3 terms of the series.. Find all x 0 such that x2 converges First we ll notice that the series converges (trivially) for x 0 since each term is identically 0. For x 0, we ll try the generalized ratio test a + a x 2(+) / ( + ) x 2 / x 2+2 ( + ) x 2 ( ) x x 2 / x 2 L 0

11 Math 352- WW08 Solutions November 9, 2008 According to the generalized ratio test, the series converges for L < and diverges for L >. So this series converges for x 2 L <, i.e., for x <. It diverges when x 2 >, i.e., when x > (since we are only ased about nonnegative values of x. The test is inconclusive at L, so we need to chec the end point x separately. (If we were considering negative values of x, too, we would also have to chec x.) When x we have x 2 2 which is a divergent p-series (p ). Therefore, the series diverges for x. To summarize, the series converges absolutely for 0 x < and diverges for x.