# Which one of the following properties does the...

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mei (dym252) – HW13 – schultz – (53990) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points To apply the root test to an infinite series summationdisplay n a n the value of ρ = lim n → ∞ | a n | 1 /n has to be determined. Compute the value of ρ for the series summationdisplay n =1 3 2 n parenleftbigg n 4 n parenrightbigg n 2 . 1. ρ = 3 e 4 2. ρ = 9 3. ρ = 4. ρ = 3 e 4 5. ρ = 9 e 4 correct Explanation: After division, n 4 n = 1 4 n , so ( a n ) 1 /n = parenleftBigg 3 2 n parenleftbigg 1 4 n parenrightbigg n 2 parenrightBigg 1 /n = 3 2 parenleftbigg 1 4 n parenrightbigg n . But lim n → ∞ parenleftbigg 1 4 n parenrightbigg n = e 4 . Consequently, ρ = 9 e 4 . 002 10.0points To apply the root test to an infinite series summationdisplay n a n the value of ρ = lim n → ∞ ( a n ) 1 /n has to be determined. Compute ρ for the series summationdisplay n =1 parenleftbigg 7 tan 1 n 6 parenrightbigg n . 1. ρ = 3 7 2. ρ = 7 12 3. ρ = 7 6 4. ρ = 7 π 12 correct 5. ρ = 3 π 7 Explanation: For the given series ( a n ) 1 /n = 7 tan 1 n 6 . But lim n → ∞ tan 1 n = tan 1 = π 2 . Consequently, ρ = 7 π 12 . 003 10.0points Determine whether the series summationdisplay n =0 ( 6) n (2 n )!
mei (dym252) – HW13 – schultz – (53990) 2 is absolutely convergent, conditionally con- vergent, or divergent. 1. conditionally convergent 2. divergent 3. absolutely convergent correct Explanation: We use the Ratio Test with a n = ( 6) n (2 n )! . For then vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle ( 6) n +1 (2 n + 2)! (2 n )! ( 6) n vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle ( 6) n +1 (2 n + 2)! (2 n )! ( 6) n vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle 6 (2 n + 1)(2 n + 2) vextendsingle vextendsingle vextendsingle . Thus lim n → ∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n → ∞ 6 (2 n + 1)(2 n + 2) = 0 < 1 . Consequently, the series is absolutely convergent . 004 10.0points Which one of the following properties does the series summationdisplay n =1 Explanation: With a n = n ( 8) n 4 n 1 we see that vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = ( n + 1)8 n +1 4 n parenleftbigg 4 n 1 n 8 n parenrightbigg = n + 1 n parenleftbigg 8 4 parenrightbigg . Thus lim n → ∞ vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = 8 4 , so by the Ratio Test the given series is divergent . 005 10.0points Which one of the following properties does the series summationdisplay n =2 ( 1) n (3 n + 2) ln n have? 1. divergent 2. absolutely convergent 3. conditionally convergent correct (3 x + 2) ln x = 3 ,
mei (dym252) – HW13 – schultz – (53990) 3 so by the Limit Comparison Test the series summationdisplay n =2 1 (3 n + 2) ln n and summationdisplay n =2 1 n ln n either both converge or both diverge. But the second series diverges by the integral test because integraldisplay 2 dx x ln x = lim t →∞ [ln(ln x )] t 2 = + . We conclude that the original series is not absolutely convergent. However it is an al- ternating series and so it may still converge conditionally. To check whether or not this is the case first we have that lim n →∞ f ( n