T3Solutions - MATH 1005A Test 3 Solutions March 6, 2009...

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MATH 1005A Test 3 Solutions March 6, 2009 Questions 1-5 are multiple choice. Circle the correct answer. Only the answer will be marked. [Marks] 1. lim n !1 [ln( n )] 2 n = [3] (a) 0 (b) 1 (c) 1 (d) 1 2 (e) Does not exist Solution: (a) 2. The sequence f r n g 1 n =0 converges for [3] (a) j r j < 1( b ) j r j c ) j r j > d ) j r e ) ¡ 1 <r 1 Solution: (e) 3. The sum of the series 1 X n =1 2 n 4 n +1 is [3] (a) 2 (b) 1 (c) 1 2 (d) 3 4 (e) 1 4 Solution: (e) 4. The series 1 X n =1 1 n p converges for [3] (a) p ¸ b ) p> c ) p d ) p< e ) 0 <p< 1 Solution: (b) 5. Which of the following series converge(s)? [3] (i) 1 X n =1 1 p n (ii) 1 X n =0 ( ¡ 1) n 2 n (iii) 1 X n =1 1 n p n (a) All (b) (i) and (ii) (c) (i) and (iii) (d) (ii) and (iii) (e) None Solution: (d) 6. Determine whether or not the series 1 X n =1 n 3 +2 n 2 +1 p n 6 converges. Justify your answer. [3] Solution: lim n !1 n 3 n 2 p n 6 =1 6 =0 ) the series diverges, by the n -th-term test.
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2 7. Determine whether or not the series 1 X n =0 3 n 4 n +1 converges. Justify your answer. [3] Solution: 3 n 4 n 3 n 4 n = μ 3 4 n ,and 1 X n =0 μ 3 4 n is convergent geometric. Thus, the series con- verges, by the comparison test. 8. Determine whether or not the series
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T3Solutions - MATH 1005A Test 3 Solutions March 6, 2009...

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