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HW14-solutions

# HW14-solutions - kim(tk5895 – HW14 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW14 – Henry – (54974) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the radius of convergence, R , of the power series ∞ summationdisplay n = 1 8 x n √ n . 1. R = ∞ 2. R = 1 correct 3. R = 0 4. R = 1 8 5. R = 8 Explanation: The given series has the form ∞ summationdisplay n =1 a n x n with a n = 8 √ n . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x , while 0 < R < ∞ (iii) if it converges when | x | < R , and (iv) diverges when | x | > R . But lim n →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n →∞ √ n √ n + 1 = 1 . By the Ratio Test, therefore, the given series (a) converges for all | x | < 1, and (b) diverges for all | x | > 1. Consequently, the given series has radius of convergence R = 1 . 002 10.0 points 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 = ( − 1 , 1) 3. converges only at x = 0 4. interval of cgce = [ − 1 , 1) correct 5. interval of cgce = [ − 1 , 1] 6. interval of cgce = [ − 4 , 4] 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 kim (tk5895) – HW14 – Henry – (54974) 2 (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....
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HW14-solutions - kim(tk5895 – HW14 – Henry –(54974 1...

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