Solutions4

# Solutions4 - 1 n 3 2 ln n ≤ 1 n 3 2 for n ≥ 3 and ∞ X n =2 1 n 3 2 converges by the p-series test since p = 3 2> 1 Thus ∞ X n =2 1 n 3 2

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MATH 2007B Test 4 Solutions November 21, 2011 [Marks] 1. For each of the following sequences, determine whether it converges or diverges. If it con- [6] verges, ±nd the limit. (a) ±² 1 2 ³ n ´ (b) { ( 2) n } (c) µ ne n Solution: (a) · · · · ² 1 2 ³ n · · · · = 1 2 n 0 the sequence converges to 0. (b) | ( 2) n | =2 n →∞ ⇒ the sequence diverges. (c) By L’H b o pital’s rule, lim x →∞ xe x = lim x →∞ x e x = lim x →∞ 1 e x =0 lim n →∞ ne n =0. Thus, the sequence converges to 0. 2. Find the sum of the series X n =0 2 · 3 n +1 4 n [4] Solution: X n =0 2 · 3 n +1 4 n = X n =0 6 ² 3 4 ³ n = 6 1 3 / 4 =24 . 3. Determine whether the series converges or diverges. Justify your answer. [20] (a) X n =0 3 n 2 n +1 Solution: X n =0 3 n 2 n +1 = X n =0 1 2 ² 3 2 ³ n diverges since it is geometric with r = 3 2 and | r |≥ 1. (b) X n =0 2 n +3 n 2 +1 Solution: lim n →∞ 2 n +3 n 2 +1 =2 6 =0 X n =0 2 n +3 n 2 +1 diverges by the n th-term test. (c) X n =2 1 n 3 / 2 ln( n

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Unformatted text preview: 1 n 3 / 2 ln( n ) ≤ 1 n 3 / 2 for n ≥ 3 and ∞ X n =2 1 n 3 / 2 converges by the p-series test since p = 3 2 > 1. Thus, ∞ X n =2 1 n 3 / 2 ln( n ) converges by the comparison test. (d) ∞ X n =2 1 √ n − 1 Solution: 1 √ n − 1 ≥ 1 √ n = 1 n 1 / 2 and ∞ X n =2 1 n 1 / 2 diverges by the p-series test since p = 1 2 ≤ 1. Hence, ∞ X n =2 1 √ n − 1 diverges by the comparison test. (e) ∞ X n =3 1 n [ln( n )] 2 Solution: f ( x ) = 1 x (ln x ) 2 is positive, continuous and decreasing for x ≥ 3, and ± ∞ 3 1 x (ln x ) 2 dx = − 1 ln x ² ² ² ² ∞ 3 = 1 ln(3) < ∞ ⇒ ∞ X n =3 1 n [ln( n )] 2 converges by the integral test....
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## This note was uploaded on 01/18/2012 for the course MATH 2007 taught by Professor S.melkonian during the Fall '11 term at Carleton CA.

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Solutions4 - 1 n 3 2 ln n ≤ 1 n 3 2 for n ≥ 3 and ∞ X n =2 1 n 3 2 converges by the p-series test since p = 3 2> 1 Thus ∞ X n =2 1 n 3 2

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