HW14-solutions - wang(kw28364 HW14 staron(53735 This...

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wang (kw28364) – HW14 – staron – (53735) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find the interval of convergence of the se- ries summationdisplay n =1 x n n + 3 . 1. converges only at x = 0 2. interval of cgce = [ 3 , 3] 3. interval of cgce = [ 1 , 1] 4. interval of cgce = ( 1 , 1) 5. interval of cgce = [ 1 , 1) correct 6. interval of cgce = ( 3 , 3] Explanation: When a n = x n n + 3 , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle x n +1 n + 4 n + 3 x n vextendsingle vextendsingle vextendsingle vextendsingle = | x | parenleftbigg n + 3 n + 4 parenrightbigg = | x | radicalbigg n + 3 n + 4 . But lim n → ∞ n + 3 n + 4 = 1 , so lim n → ∞ vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = | x | . By the Ratio Test, therefore, the given series (i) converges when | x | < 1, (ii) diverges when | x | > 1. We have still to check what happens at the endpoints x = ± 1. At x = 1 the series becomes ( ) summationdisplay n =1 1 n + 3 . Applying the Integral Test with f ( x ) = 1 x + 3 we see that f is continuous, positive, and de- creasing on [1 , ), but the improper integral I = integraldisplay 1 f ( x ) dx diverges, so the infinite series ( ) diverges also. On the other hand, at x = 1, the series becomes ( ) summationdisplay n =1 ( 1) n n + 3 . which is an alternating series summationdisplay n =1 ( 1) n a n , a n = f ( n ) with f ( x ) = 1 x + 3 the same continuous, positive and decreasing function as before. Since lim x → ∞ f ( x ) = lim x → ∞ 1 x + 3 = 0 , however, the Alternating Series Test ensures that ( ) converges. Consequently, the interval of convergence = [ 1 , 1) . 002 10.0points Determine the radius of convergence, R , of the series summationdisplay n =1 ( 6) n n + 3 x n .
003 10.0points
wang (kw28364) – HW14 – staron – (53735) 3 But as n → ∞ , neither of 2 n, ( 1) n 2 n −→ 0 holds, so both of summationdisplay n =1 2 n, summationdisplay n =1 ( 1) n 2 n diverge. Consequently, the interval of convergence = ( 1 , 1) .

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