HW14 Solutions.pdf - farr(arf2325 u2013 HW14 u2013 meth...

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farr (arf2325) – HW14 – meth – (54160) 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 + 4 . 1. interval of cgce = ( 4 , 4] 2. interval of cgce = [ 4 , 4] 3. converges only at x = 0 4. interval of cgce = [ 1 , 1) correct 5. interval of cgce = [ 1 , 1] 6. interval of cgce = ( 1 , 1) Explanation: When a n = x n n + 4 , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle x n +1 n + 5 n + 4 x n vextendsingle vextendsingle vextendsingle vextendsingle = | x | parenleftbigg n + 4 n + 5 parenrightbigg = | x | radicalbigg n + 4 n + 5 . But lim n → ∞ n + 4 n + 5 = 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 + 4 . Applying the Integral Test with f ( x ) = 1 x + 4 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 + 4 . which is an alternating series summationdisplay n =1 ( 1) n a n , a n = f ( n ) with f ( x ) = 1 x + 4 the same continuous, positive and decreasing function as before. Since lim x → ∞ f ( x ) = lim x → ∞ 1 x + 4 = 0 , however, the Alternating Series Test ensures that ( ) converges. Consequently, the interval of convergence = [ 1 , 1) . 002 10.0points Find the interval of convergence of the se- ries summationdisplay n =1 ( 1) n x n 3 n 2 + 5 .
farr (arf2325) – HW14 – meth – (54160) 2 1. [ 1 , 1) 2. converges only at x = 0 3. ( 1 , 1] 4. [ 3 , 5] 5. [ 1 , 1] correct 6. ( 3 , 5] Explanation: When a n = ( 1) n x n 3 n 2 + 5 , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle x n +1 3( n + 1) 2 + 5 3 n 2 + 5 x n vextendsingle vextendsingle vextendsingle vextendsingle = | x | parenleftbigg 3 n 2 + 5 3( n + 1) 2 + 5 parenrightbigg . But 3( n + 1) 2 + 5 = 3 n 2 + 6 n + 8 , while lim n → ∞ 3 n 2 + 5 3 n 2 + 6 n + 8 = 1 . Thus 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 3 n 2 + 5 . Applying the Integral Test with f ( x ) = 1 3 x 2 + 5 we see that f is continuous, positive, and de- creasing on [1 , ); in addition,the improper integral I = integraldisplay 1 f ( x ) dx converges, so the infinite series ( ) converges also. On the other hand, at x = 1, the series becomes ( ) summationdisplay n =1 ( 1) n 3 n 2 + 5 .

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