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Math 1280-1 Peter Alfeld. Webwork Problem Set 1. Due 2/7/06 at 11:59 PM. Procrastination Is Hazardous!

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Math 1280-1 Peter Alfeld. Webwork Problem Set 1. Due 2/7/06 at 11:59 PM. Procrastination Is Hazardous! n! 7. n1000 Math 1280-1 n3 5n 8. 3n−n5 − Peter Alfeld. 4.(1 pt) Determine whether the sequences are in- creasing, decreasing, or not monotonic. If increasing, enter 1 as your answer. If decreasing, enter 1as your WeBWorK Problem Set 1. answer. If not monotonic, enter 0 as your answer− . 1 1. an = 2n+6 Due 2/7/06 at 11:59 PM. n 2 2. an = n+−2 Get to work on this set right away and an- p + 3. a = n 2 • swer these questions well before the deadline. n 6n+2 4. a = cosn Not only will this give you the chance to fig- n 2n ure out what's wrong if an answer is not ac- 5.(1 pt) Consider the sequence cepted, you also will avoid the likely rush and ncos(np) congestion prior to the deadline. a = : n 2n 1 − Write the first five terms of an, and find limn ∞ an. Procrastination is hazardous! If the sequence diverges, enter ”divergent” in the! an- 1.(1 pt) Determine whether the sequence is diver- swer box for its limit. gent or convergent. If it is convergent, evaluate its a) First five terms: , , , , . limit. If it diverges to infinity, state your answer as b) limn ∞ an = . ! ”INF” (without the quotation marks). If it diverges 6.(1 pt) Consider the sequence to negative infinity, state your answer as ”MINF”. If it diverges without being infinity or negative infinity, ln(1=n) an = : state your answer as ”DIV”. p2n 4n7 + sin2(11n) lim Write the first five terms of an, and find limn ∞ an. n ∞ 6 + ! ! n 10 If the sequence diverges, enter ”divergent” in the an- swer box for its limit. 2.(1 pt) Find the limit of the sequence whose terms a) First five terms: , , , , . are given by b) limn ∞ an = . a = (n2)(1 cos( 5:5 )): ! n − n 7.(1 pt) Suppose (1 pt) Match each sequence below to statement 3. 1 2 3 4 5 that BEST fits it. a1 = ;a2 = ;a3 = ;a4 = ;a5 = : 2 1 3 1 4 1 5 1 6 1 STATEMENTS − 2 − 3 − 4 − 5 − 6 Z. The sequence converges to zero; a) Find an explicit formula for an: I. The sequence diverges to infinity; . F. The sequence has a finite non-zero limit; b) Determine whether the sequence is convergent D. The sequence diverges. or divergent: . SEQUENCES (Enter ”convergent” or ”divergent” as appropriate.) ( 1 ) c) If it converges, find limn ∞ an = 1. nsin n ! (ln(n)) . 2. n 8.(1 pt) Suppose n100 3. (1:01)n 4. ln(ln(ln(n))) 1 2 a1 = 2;an+1 = an + : 5. sin(n) 2 an 6. arctan(n + 1) Find limn ∞ an = . 1 ! ∞ Hint: Let a = limn ∞. Then, since an+1 = 7 ! 5. 1 2 1 2 ∑ n10 16 an + , we have a = a + . Now solve n=1 − 2 an 2 a for a. 11.(1 pt) Determine the sum of the following se- ∞ 7 9.(1 pt) Consider the series ∑n=1 . Let sn be the ries. n+6 ∞ n 1 n-th partial sum; that is, ( 4) − n ∑ − n 7 n= 6 sn = ∑ : 1 i=1 i + 6 (1 pt) Determine the sum of the following se- Find s4 and s8 12. ries. s4 = ∞ 4n + 6n s8 = ∑ ( n ) 10.(1 pt) Match each of the following with the cor- n=1 11 rect statement. C stands for Convergent, D stands for Divergent. 13.(1 pt) Express 8:846846846::: as a rational p ∞ 1 number, in the form q 1. ∑ 5 where p and q have no common factors. 6 + pn3 n=1 p = and ∞ 7 2. q = ∑ n(n + 7) n=1 14.(1 pt) A ball drops from a height of 18 feet. ∞ 2 + 7n Each time it hits the ground, it bounces up 30 per- 3. ∑ n n=1 8 + 9 cents of the height it fall. Assume it goes on forever, ∞ ln(n) 4. ∑ find the total distance it travels. n=1 10n Prepared by the WeBWorK group, Dept. of Mathematics, University of Rochester, c UR 2 A. convergent Math 1280-1 • B. divergent • 5.(1 pt) Determine whether the following series is Peter Alfeld. ∞ 1 ( )n+1 ∑ 1 1:1 WeBWorK Problem Set 2. n=1 − 5n A. conditionally convergent Due 2/7/06 at 11:59 PM. • B. absolutely convergent • Get to work on this set right away and an- C. divergent • • swer these questions well before the deadline. 6.(1 pt) Determine whether the following series is Not only will this give you the chance to fig- ure out what's wrong if an answer is not ac- ∞ 1 ( 1)n cepted, you also will avoid the likely rush and ∑ − pn2 1 n=2 − congestion prior to the deadline. A. absolutely convergent • B. conditionally convergent Procrastination is hazardous! • C. divergent (1 pt) Consider the series: • 1. 7.(1 pt) Each of the following statements is an at- ∞ 3 3 tempt to show that a given series is convergent or not using the Comparison Test (NOT the Limit Compari- ∑ (k 1)2 − k2 k=10 − son Test.) For each statement, enter C (for ”correct”) a) Determine whether the series is convergent or if the argument is valid, or enter I (for ”incorrect”) divergent: . if any part of the argument is flawed. (Note: if the (Enter ”convergent” or ”divergent” as appropriate.) conclusion is true but the argument that led to it was b) If it converges, find its sum: . wrong, you must enter I.) If the series diverges, enter here ”divergent” again. ln(n) 1. For all n > 2, > 1 , and the series ∑ 1 (1 pt) Determine the convergence or divergence n2 n2 n2 2. converges, so by the Comparison Test, the se- of the following series. ln(n) ries ∑ n2 converges. ∞ p 1 2 1 2n + 1 2. For all n > 1, nln(n) < n , and the series 2∑ n ∑ 2 n=1 n diverges, so by the Comparison Test, the se- 1 A. convergent ries ∑ nln(n) diverges. • B. divergent n > ln(n) < 1 1 • 3. For all 1, n2 n1:5 , and the series ∑ n1:5 3.(1 pt) Determine the convergence or divergence converges, so by the Comparison Test, the se- of the following series. ln(n) ries ∑ n2 converges. 4. For all n > 2, n < 2 , and the series 2∑ 1 ∞ 3k + k n3 7 n2 n2 ∑ converges, so by−the Comparison Test, the se- n=1 k! n ries ∑ n3 7 converges. A. convergent − • 5. For all n > 1, arctan(n) < π , and the series B. divergent n3 2n3 • π 1 2 ∑ n3 converges, so by the Comparison Test, 4.(1 pt) Determine the convergence or divergence arctan(n) of the following series. the series ∑ n3 converges. ln(n) 6. For all n > 2, > 1 , and the series ∑ 1 di- ∞ 1 n n n pn 1 cos verges, so by the Comparison Test, the series ∑ ln(n) n=1 − n ∑ n diverges. 1 ∞ 1 6. 8.(1 pt) The three series ∑An, ∑Bn, and ∑Cn have ∑ n=1 pn terms 1 1 1 10.(1 pt) Find the convergence set of the given A = ; B = ; C = : n n7 n n4 n n power series: Use the Limit Comparison Test to compare the fol- ∞ 2nxn lowing series to any of the above series. For each of ∑ the series below, you must enter two letters. The first n=1 n! is the letter (A,B, or C) of the series above that it can The above series converges for < x < . be legally compared to with the Limit Comparison Enter ”infinity” for ¥ and ”-infinity” for ¥. Test. The second is C if the given series converges, − 11.(1 pt) A famous sequence fn, called the Fi- or D if it diverges. So for instance, if you believe bonacci Sequence after Leonardo Fibonacci, who in- the series converges and can be compared with series troduced it around A.D. 1200, is defined by the recur- C above, you would enter CC; or if you believe it di- sion formula verges and can be compared with series A, you would enter AD. f1 = f2 = 1; fn+2 = fn+1 + fn: ∞ 5n2 + 5n6 1. Find the radius of convergence of ∑ 4n7 + 6n3 4 n=1 − ∞ 4n4 + n7 ∞ 2. f xn: ∑ 187n11 + 6n4 + 5 ∑ n n=1 n=1 ∞ 5n4 + n2 5n 3. − Radius of convergence: . ∑ 6n11 5n9 + 1 n=1 − 12.(1 pt) Find the interval of convergence for the given power series. 9.(1 pt) Select the FIRST correct reason why the ∞ n4(x + 7)n given series diverges. ∑ 14 ( n)(n 3 ) A. Diverges because the terms don't have limit n=1 10 zero The series is convergent: B. Divergent geometric series from x = , left end included (Y,N): C. Divergent p series to x = , right end included (Y,N): D. Integral test 13.(1 pt) Match each of the power series with its E. Comparison with a divergent p series interval of convergence. F. Diverges by limit comparison test ∞ (x 5)n G. Cannot apply any test done so far in class 1. − ∑ (n!)5n ∞ n=1 cos(np) ∞ n!(5x 5)n 1. ∑ 2. − n=1 ln(5) ∑ 5n ∞ n=1 1 ∞ (5x)n 2. ∑ 3. n=1 nln(n) ∑ n5 ∞ 2 n n=1 (n + 1)(6 + 1) ∞ (x 5)n 3. ∑ 2n 4. − n=1 6 ∑ (5)n ∞ n=1 n (2n)! 1 1 4. ( 1) A. [ − ; ] ∑ − (n!)2 5 5 n=1 B. 5=5 ∞ 7n + 4 C. (f ¥;g¥) 5. ∑ n − n=1 ( 1) D.
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