11.5 Alternating Series Test-Absolute Convergence-solutions

11.5 Alternating Series Test-Absolute Convergence-solutions...

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im (gi768) – 11.5 Alternating Series Test/Absolute Convergence – sadun – (53088) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Problem 2 is a little tricky, since it’s not EXACTLY alternating. You have to do some manipulations before you can apply any of our standard tests. 001 10.0points Which one of the following series is conver- gent? 1. summationdisplay n =1 ( 1) n 1 2 + n correct 2. summationdisplay n =1 ( 1) n 1 2 + n 4 + n 3. summationdisplay n =1 3 4 + n 4. summationdisplay n =1 ( 1) 2 n 4 3 + n Explanation: Since summationdisplay n =1 ( 1) 2 n 4 3 + n = summationdisplay n =1 4 3 + n , use of the Limit Comparison and p -series Tests with p = 1 / 2 shows that this series is divergent. The same use of the Limit Comparison and p -series Tests with p = 1 / 2 shows that the series summationdisplay n =1 3 4 + n also is divergent. On the other hand, by the Divergence Test, the series summationdisplay n =1 ( 1) n 1 2 + n 4 + n is divergent because lim n → ∞ ( 1) n 1 2 + n 4 + n negationslash = 0 . This leaves only the series summationdisplay n =1 ( 1) n 1 2 + n . To see that this series is convergent, set b n = 1 2 + n . Then (i) b n +1 b n , (ii) lim n → ∞ b n = 0 . Consequently, by the Alternating Series Test, the series summationdisplay n =1 ( 1) n 1 2 + n is convergent. 002 10.0points Determine if the series summationdisplay n =0 1 n + 1 cos parenleftBig 4 parenrightBig converges or diverges. Similarseriesarefundamentaltothemath- ematicaldescriptionofallmusicalsounds ! 1. series converges correct 2. series diverges Explanation: The basic idea is that cos parenleftBig π 4 x parenrightBig is a peri- odic function taking values between 1 and 1. This will enable us to write the given series as a sum of alternating series. Indeed, cos(0) = 1 , cos parenleftBig π 4 parenrightBig = 1 2 , cos parenleftBig π 2 parenrightBig = 0 , cos parenleftbigg 3 π 4 parenrightbigg = 1 2 ,
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im (gi768) – 11.5 Alternating Series Test/Absolute Convergence – sadun – (53088) 2 while cos( π ) = 1 , cos parenleftbigg 5 π 4 parenrightbigg = 1 2 , cos parenleftbigg 3 π 2 parenrightbigg = 0 , cos parenleftbigg 7 π 4 parenrightbigg = 1 2 , and so on for higher multiples of π 4 . Thus the given series can be written as parenleftBig 1 + 1 2 2 1 4 2 1 5 1 6 2 + 1 8 2 + 1 9 1 10 2 1 12 2 1 13 . . .
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