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**Unformatted text preview: **004 10.0 points To apply the root test to an infinite series ∑ 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 = 9 e − 4 correct 4. ρ = ∞ 5. ρ = 3 e 4 Explanation: momin (rrm497) – Homework 12 – cheng – (58520) 3 After division, n − 4 n = 1 − 4 n , so ( a n ) 1 /n = parenleftBigg 3 2 n braceleftbigg 1 − 4 n bracerightbigg 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 . 005 10.0 points To apply the ratio test to the infinite series summationdisplay n a n , the value λ = lim n →∞ a n +1 a n has to be determined. Compute λ for the series ∞ summationdisplay n =1 2 n 4 n 2 + 5 . 1. λ = 2 correct 2. λ = 2 5 3. λ = 1 2 4. λ = 0 5. λ = 2 9 Explanation: By algebra, a n +1 a n = 2 n +1 2 n bracketleftbigg 4 n 2 + 5 4( n + 1) 2 + 5 bracketrightbigg . But 4 n 2 + 5 4( n + 1) 2 + 5 = 4 + 5 n 2 4 parenleftbigg n + 1 n parenrightbigg 2 + 5 n 2 . Since parenleftbigg n + 1 n parenrightbigg 2 = parenleftbigg 1 + 1 n parenrightbigg 2 → 1 as n → ∞ , we see that lim n →∞ a n +1 a n = lim n →∞ 2 parenleftBig 4 + 5 n 2 parenrightBig 4 parenleftBig n + 1 n parenrightBig 2 + 5 n 2 = 2 . Consequently, λ = 2 . 006 10.0 points Which one of the following properties does the series ∞ summationdisplay n = 1 ( − 1) n − 1 5 n 2 + 3 2 n have? 1. absolutely convergent correct 2. divergent 3. conditionally convergent Explanation: The given series has the form ∞ summationdisplay n =1 ( − 1) n − 1 b n , b n = 5 n 2 + 3 2 n momin (rrm497) – Homework 12 – cheng – (58520) 4 of an alternating series. But the denominator is increasing very fast, so first let’s check if the series is absolutely convergent rather than simply conditionally convergent. We use the Ratio test, for then vextendsingle vextendsingle vextendsingle ( − 1) n b n +1 ( − 1) n − 1 b n vextendsingle vextendsingle vextendsingle = b n +1 b n = 1 2 5( n + 1) 2 + 3 5 n 2 + 3 . But 5( n + 1) 2 + 3 5 n 2 + 3 = 5 n 2 + 10 n + 8 5 n 2 + 3 −→ 1 as n → ∞ . Thus lim n →∞ vextendsingle vextendsingle vextendsingle ( − 1) n b n +1 ( − 1) n − 1 b n vextendsingle vextendsingle vextendsingle = 1 2 < 1 . Consequently, by the Ratio test, the given series is absolutely convergent . 007 10.0 points Determine whether the series ∞ summationdisplay n = 1 ( − 1) n +8 √ n is absolutely convergent, conditionally con- vergent, or divergent. 1. conditionally convergent correct 2. divergent 3. absolutely convergent Explanation: By the Alternating Series test, the series ∞ summationdisplay n = 1 ( − 1) n +8 √ n converges. On the other hand, by the p-series test with p = 1 2 ≤ 1, the series ∞ summationdisplay n =1 1 √ n is divergent. Consequently, the series is conditionally convergent ....

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