11.6 Ratio and Root Test-solutions

11.6 Ratio and Root Test-solutions - im(gi768 11.6 Ratio...

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im (gi768) – 11.6 Ratio and Root Test – sadun – (53088) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine whether the series s n =1 ( - 1) n 1 e 1 /n 4 n is absolutely convergent, conditionally con- vergent or divergent. 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: Since s n =1 ( - 1) n 1 e 1 /n 4 n = - 1 4 s n = 1 ( - 1) n e 1 /n n , we have to decide iF the series s n =1 ( - 1) n e 1 /n n is absolutely convergent, conditionally con- vergent, or divergent. ±irst we check For absolute convergence. Now, since e 1 /n 1 For all n 1, e 1 /n 4 n 1 4 n > 0 . But by the p -series test with p = 1, the series s n = 1 1 4 n diverges, and so by the Comparison Test, the series s n = 1 e 1 /n 4 n too diverges; in other words, the given series is not absolutely convergent. To check For conditional convergence, con- sider the series s n =1 ( - 1) n f ( n ) where f ( x ) = e 1 /x 4 x . Then f ( x ) > 0 on (0 , ). On the other hand, f ( x ) = - 1 4 x 3 e 1 /x - e 1 /x 4 x 2 = - e 1 /x p 1 + x 4 x 3 P . Thus f ( x ) < 0 on (0 , ), so f ( n ) > f ( n + 1) For all n . ±inally, since lim x →∞ e 1 /x = 1 , we see that f ( n ) 0 as n → ∞ . By the Alternating Series Test, thereFore, the series s n =1 ( - 1) n f ( n ) is convergent. Consequently, the given series is conditionally convergent . keywords: 002 10.0 points Determine whether the series s k =1 ( - 1) k 1 cos 2 ( k ) 2 k is absolutely convergent, conditionally con- vergent or divergent 1. absolutely convergent correct 2. conditionally convergent 3. divergent Explanation:
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im (gi768) – 11.6 Ratio and Root Test – sadun – (53088) 2 To check for absolute convergence we have to decide if the series s k =1 cos 2 ( k ) 2 k is convergent. For this we can use the Com- parison Test with a k = cos 2 ( k ) 2 k , b k = 1 2 k .
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