FINAL REVIEW #2-solutions

FINAL REVIEW #2-solutions - suleimenov(bs26835 – FINAL...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: suleimenov (bs26835) – FINAL REVIEW #2 – rusin – (55565) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. More of the final review. 001 10.0 points Find the interval of convergence of the se- ries ∞ summationdisplay n =1 x n √ n + 4 . 1. converges only at x = 0 2. interval of cgce = [ − 1 , 1] 3. interval of cgce = [ − 1 , 1) correct 4. interval of cgce = ( − 4 , 4] 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 (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 10.0 points suleimenov (bs26835) – FINAL REVIEW #2 – rusin – (55565) 2 Determine the interval of convergence of the series ∞ summationdisplay k = 1 ( − 1) k − 1 1 k 2 ( x − 1) k . 1. interval of convergence = [ − 2 , 0 ) 2. interval of convergence = (0 , 2 ) 3. interval of convergence = [0 , 2 ] cor- rect 4. interval of convergence = (0 , 2 ] 5. interval of convergence = [0 , 2 ) 6. interval of convergence = ( − 2 , 0 ] 7. interval of convergence = [ − 2 , 0 ] 8. interval of convergence = ( − 2 , 0 ) Explanation: The given series has the form ∞ summationdisplay k = 1 c k ( x − 1 ) k , c k = ( − 1) k − 1 k 2 ....
View Full Document

{[ snackBarMessage ]}

Page1 / 12

FINAL REVIEW #2-solutions - suleimenov(bs26835 – FINAL...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online