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Unformatted text preview: NOTE: PRACTICE COPY #1 to be done in class (no other answers provided.) ANSWERS  EXAM #1 PRACTICE COPY #2 To be done in the review session on SATURDAY, Feb 11, 2007 from 7 9pm in C106 Holmes Hall.
1. 4. 4 7 2. 72 67 3. 1.0746 a) Since n starts at 1, everything is positive since there are no negatives anywhere in the problem. Since the denominator is only zero when n = 1, it is continuous everywhere except at n = 1, but n starts at 1 so it is continuous on our interval! b) Show an +1 < an OR Show f ' is negative beyond some value of n c) DIVERGES 5. DIVERGES 6. CONVERGES 7. 10. CONVERGES DIVERGE . 8. CONVERGES 9. CONVERGES . ANSWERS  EXAM #1 PRACTICE COPY #3 To be done in the review session on SATURDAY, Feb 11, 2007 from 7 9pm in C106 Holmes Hall.
1. 35 24 2. 4 19 3. 0.21 4. a) b) c) Since n starts at 1 and goes to infinity and there are no negatives anywhere in the problem, this is a POSITIVE function. Also, the denominator can never be negative, so it is CONTINUOUS everywhere! Show an +1 < an OR Show f ' is negative beyond some value of n CONVERGE . 6. 9. CONVERGE . CONVERGES . 7. 10. DIVERGES DIVERGE. 5. 8. DIVERGES DIVERGES DETAILED ANSWERS  EXAM #1 PRACTICE COPY #4
1. First part is geometric: a= 4 1 r = <1 25 5 4 4 5 25 1 = = 4 25 4 5 5 4 4 4 + + +... 52 53 54 4 a 25 Converges to = = 1r 1 1  5 Second part is telescoping: 2 A B = + ( n + 1) (n + 2) n + 1 n + 2 2 = A(n + 2) + B (n + 1) if n = 2 : 2 =  B B = 2 if n = 1 : 2 = A A=2 n +1
n=2 7 2  1 2 1 = 2 n +1  n + 2 n+2 n= 2 Goes to zero as n 7 1 1 1 1  1  1  1  1 ... + 1 sn = 2 + + + + 3 4 +1 4 5 5 6 6 7 n 1 2 lim sn = 2 = n7 3 3 Subtract the two answers to get the final answer: 2.  } 1 n+2 1 2 3 10 7  =  = 5 3 15 15 15 ( 1) = 1 + 1  1 + ... 2n 5 (3!) n =1 5 ( n!) 5 (1!) 5 (2!) 123
7 n
2 4 6 .00001 The third term is the ERROR term. Therefore, I only need to add the first TWO terms: =  .0392 3 a)
7 n =1 n 2 e  n Show :
2 ( n + 1) 2 e  ( n +1) < n 2 e  n ( n + 1)
e n +1 n2 < n e e n < n 2 e n +1 True for n 2. ( n + 1) 2 n 2 + 2n + 1 < n 2 e 2n + 1 < (e  1)n 2 Or, you can use the derivative and show it is negative beyond a certain point: f ( x) = x 2 e  x f ( x) = x 2 (e  x ) + e  x (2 x ) = e  x ( x 2 + 2 x) = The terms are decreasing beyond x = 2. b)
7  x( x  2) 7 ex neg for x > 2 pos x e
2 1 x dx = lim e  x dx x2
b 1 b Tabular Alt sign +  +  u x2 2x 2 0 dv e x  e x e x  e x
b lim (  x e
b 2 x  2 xe x  2e x b 1 ) x2 2 x 2 = lim x  x  x  b e e e 1 b 2 2b 2 1 2 2 5 lim b  b  b      = b e e 4 4 2 4 4e4 e e e e 1 4 3
goes to zero . Both the integral and the series CONVERGE . 4. A.S .T . 1. 2. lim n =0 n7 n +4 n +1 n Show : < 2 2 ( n + 1) + 4 n + 4
2 ( n + 1) ( n 2 + 4 ) < n ( n + 1) + 4
2 n3 + 4n + n 2 + 4 < n ( n 2 + 2n + 1 + 4 ) n 3 + 4 n + n 2 + 4 < n 3 + 2 n 2 + 5n 4 < n2 + n True for n 2. ( ) 7 5. 6. 7. Re write :
n =1 1 n10 / 3 p  series p = 10 / 3 > 1 CONVERGES
n 6 6n  1 n 1 ROOT TEST : lim n = 3 >1 = lim n7 2n 2n n Compare to DIVERGES 1n 1 HARMONIC DIVERGES P n 3 + 2n  5 4 5n 4 + 2 n 3 + 2n  5 n n 4 + 2n 2  5n 4 4 4 lim = lim = lim = n 4 4 f n' a n 1 5n + 2 1 5n + 2 5 n They grow at the same rate, so both DIVERGE. 8. 1 n = 1 n ln n
Integral is easy. This is positive, continuous, and decreasing. We can use the integral test. 1 1 du ln x dx u = ln x du = x dx = ln u + C = ln ln x + C x u b 1 b lim 1 dx = lim ln ln x 1 = lim ln ln b  ln ln1 = { b' fa b b 123 x ln x 0 1 123 goes to 7
goes to  P Both the integral and the series DIVERGE . 9. a) FALSE. It has to grow FASTER than a diverging series or SLOWER than a CONVERGING series or we cannot tell what happens! b) TRUE. ANSWERS  MORE PRACTICE: These are only BRIEF answers. No work is shown! You should show all work on the exam!
1. 4. 3 5 2 2. 5. 12 1 10 3. 6. 1 2 2 9. CONVERGES 12. DIVERGES 15. CONVERGES 18. DIVERGES 21. CONVERGES 24. CONVERGES 27. 0.2141 7. DIVERGES 10. DIVERGES 13. CONVERGES 16. CONVERGES 19. CONVERGES 22. CONVERGES 25. 0.2231 or 0.2232 8. DIVERGES 11. CONVERGES 14. DIVERGES 17. CONVERGES 20. CONVERGES 23. CONVERGES 26. 0.3934 or 0.3935 Also, do the homework in Section 11.7 for more practice. In this section, all of the series are mixed together. It will give you an idea of whether or not you know what tests to use when! ...
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This note was uploaded on 04/10/2008 for the course LBS 119 taught by Professor Hanninichols during the Spring '08 term at Michigan State University.
 Spring '08
 HanniNichols

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