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# N 1 n 1 n 1 n n parenrightbigg now n n 1 1 n 1 while

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Unformatted text preview: ( n + 1)! ( n + 1) n +1 n n parenrightbigg . Now n ! ( n + 1)! = 1 n + 1 , while ( n + 1) n +1 n n = ( n + 1) parenleftbigg n + 1 n parenrightbigg n . Thus vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 4 parenleftbigg n + 1 n parenrightbigg n −→ 4 e > 1 as n → ∞ , so series ( A ) diverges. ( B ) The root test is the better one to apply because | a n | 1 /n = 3 n + 5 −→ as n → ∞ , so series ( B ) converges. Consequently, of the given infinite series, only A diverges . 013 10.0 points Determine which, if any, of the series A. ∞ summationdisplay m = 3 m + 3 m 2 ln m + 2 B. 3 5 + 4 6 + 5 7 + 6 8 + 7 9 + . . . are convergent. 1. neither of them correct 2. both of them 3. B only 4. A only Explanation: momin (rrm497) – Homework 12 – cheng – (58520) 8 A. Divergent: use Limit Comparison Test and Integral Test with f ( x ) = 1 x ln x . B. Divergent by Divergent Series Test: given series is of the form ∞ summationdisplay n = 3 n n + 2 , and so lim n →∞ a n = 1 negationslash = 0 . 014 10.0 points Which, if any, of the following statements are true? A. The Ratio Test can be used to deter- mine whether the series summationdisplay n = ∞ 1 /n ! converges or diverges. B. The Root Test can be used to determine whether the series ∞ summationdisplay k = 1 parenleftBig ln k 2 + k parenrightBig k converges or diverges. 1. both of them correct 2. A only 3. B only 4. neither of them Explanation: A. True: when a n = 1 /n !, then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = 1 n + 1 −→ as n → ∞ , so ∑ a n is convergent by Ratio Test. B. True: when a k = parenleftBig ln k 2 + k parenrightBig k , then | a k | 1 /k = ln k 2 + k −→ as k → ∞ , so ∑ a k is convergent by the Root Test. 015 10.0 points Determine which, if any, of the series A. ∞ summationdisplay n = 2 3 √ n n √ n − 2 B. ∞ summationdisplay k = 1 k + 2 k 3 k are convergent. 1. A only 2. both of them 3. neither of them 4. B only correct Explanation: A. Divergent: use Limit Comparison Test and p-series Test with p = 1. B. Convergent: use Limit Comparison Test and Geometric Series. 016 10.0 points Determine whether the series ∞ summationdisplay k = 1 ( − 1) k − 1 cos parenleftBig 1 4 k parenrightBig is absolutely convergent, conditionally con- vergent or divergent. momin (rrm497) – Homework 12 – cheng – (58520) 9 1. absolutely convergent 2. series is divergent correct 3. conditionally convergent Explanation: The given series can be written as ∞ summationdisplay k = 1 ( − 1) k − 1 a k with a k = cos parenleftBig 1 4 k parenrightBig > , and so is an alternating series. But lim k →∞ cos parenleftBig 1 4 k parenrightBig = 1 , in which case lim k →∞ ( − 1) k − 1 a k does not exist. Thus by the Divergence test, the given series is divergent ....
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