test4a

# test4a - MA 241 Test 4 Form A-No calculators No cell phones...

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Unformatted text preview: MA 241 Test 4 Form A -No calculators, No cell phones, No PDA's -No credit for guessing -Please do not ask questions during the test unless you think there is a typing error. -Please begin each problem on a new page. I.(80pts) Determine if the series converges or diverges. Name the test, show the criteria for the test is met and make a concluding statement. Be sure to "X" out any work you do not want graded. Omit one problem of the 6 problems. Work 5 problems of your choice. Indicate the omitted problem, otherwise the first 5 will be graded. 2 2n-1 1. n n=1 3 1 2. 2 n=1 1+ n 3. n - l n(n) n=1 1 n 4. n=1 3n + 2 (-1) n 3n 5. 2 n=1 n 6. n=1 (-1) n 2n n! II. (20pts) Find the interval convergence for the power series. Be sure to check the endpoints for convergence. xn n=1 n Bonus: (8pts) No partial credit. Do not attempt this until you have finished your test. Find all values of x for which the given series is a convergent geometric series. Then express the sum of the series as a function of x. n (3x) n=1 MA 241 Spring 2007 Test 4 solution key Blue version Determine if the series converges or diverges. Name the test, show the criteria for the test are met and make a concluding statement. Problem 2 1 1 + n2 n=1 Solution: easiest test: comparison (other possibilities: integral test, limit comparison test). 1 1 1 1 an = 1+n2 < n2 . Let bn = n2 . We know n2 converges because it is a p-series with n=1 p = 2 1. Since bn converges and it is bigger than our original series an , then n=1 n=1 an must also converge. n=1 Problem 4 (-1)n n=1 n 3n + 2 Solution: easiest test: nth term test. n 1 limn (-1)n 3n+2 = limn (-1)n 3+2/n = 0. Since the terms are not going to 0, the series does not converge. Problem 5 n=1 3n n2 Solution: easiest test: ratio test (other possibilities: nth term test, comparison test). an+1 | = an = = = = n2 3n+1 n| n (n + 1)2 3 n 3 3 n2 lim | n| n (n + 1)2 3 n2 3 | lim | n (n + 1)2 1 n 2 lim |3 ( )| n n+1 3 lim | n lim | Since 3 > 1, we conclude that the series must diverge. Problem 6 n=1 2n n! Solution: easiest test: ratio test (other possibilities: alternating series test). 1 n lim | an+1 | = an = = = = 2n+1 n! n| n (n + 1)! 2 n 2 2 n! lim | n| n (n + 1) (n!) 2 1 2 | lim | n (n + 1) 1 2 lim | | n (n + 1) 0 lim | Since 0 < 1, we conclude that the series must converge. 2 ...
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## This note was uploaded on 03/20/2008 for the course MA 241 taught by Professor Mccollum during the Spring '08 term at N.C. State.

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