# 003 100 points to apply the ratio test to the

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Unformatted text preview: 003 10.0 points To apply the ratio test to the infinite series summationdisplay n a n , the value λ = lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle has to be determined. Compute λ for the series ∞ summationdisplay n =1 7 n 4 n 7 + 2 . 1. λ = 7 4 2. λ = 7 6 chavez (slc2554) – 11.6: Root and Ratio Tests – meth – (91825) 2 3. λ = 0 4. λ = 7 2 5. λ = 7 correct Explanation: By algebra, a n +1 a n = 7 n +1 7 n parenleftbigg 4 n 7 + 2 4( n + 1) 7 + 2 parenrightbigg . But 4 n 7 + 2 4( n + 1) 7 + 2 = 4 + 2 n 7 4 parenleftbigg n + 1 n parenrightbigg 7 + 2 n 7 . Since parenleftbigg n + 1 n parenrightbigg 7 = parenleftbigg 1 + 1 n parenrightbigg 7 → 1 as n → ∞ , we see that lim n →∞ a n +1 a n = lim n →∞ 7 parenleftBig 4 + 2 n 7 parenrightBig 4 parenleftBig n + 1 n parenrightBig 7 + 2 n 7 = 7 . Consequently, λ = 7 . 004 10.0 points Determine whether the series ∞ summationdisplay n =1 (- 1) n- 1 sin 2 ( n ) 2 n is absolutely convergent, conditionally con- vergent or divergent 1. divergent 2. conditionally convergent 3. absolutely convergent correct Explanation: To check for absolute convergence we have to decide if the series ∞ summationdisplay n = 1 sin 2 ( n ) 2 n is convergent. For this we can use the Com- parison Test with a n = sin 2 ( n ) 2 n , b n = 1 2 n . For then ≤ a n ≤ b n , since 0 ≤ sin 2 ( n ) ≤ 1. Thus the series ∞ summationdisplay n = 1 sin 2 ( n ) 2 n converges if the series ∞ summationdisplay n = 1 1 2 n converges. But this last series is a geometric series with r = 1 2 < 1 , hence convergent. Consequently, the given series is absolutely convergent . 005 10.0 points Determine whether the following series ∞ summationdisplay n =1 (- n ) n 5 2 n +1 is absolutely convergent, conditionally con- vergent, or divergent. chavez (slc2554) – 11.6: Root and Ratio Tests – meth – (91825) 3 1. absolutely convergent 2. conditionally convergent 3. divergent correct Explanation: Since the n th term of the series has the form a n = (- 1) n 5 parenleftBig n 25 parenrightBig n , we use the Root Test, for then | a n | 1 /n = 1 5 1 /n parenleftBig n 25 parenrightBig → ∞ as n → ∞ . The Root Test thus ensures that the series is divergent ....
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