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Unformatted text preview: CALCULUS 153: SAMPLE FINAL SOLUTIONS Problem 1 (40 points) . Determine whether the following series converge or diverge. Show your work and make sure you state what test you are using. (10 points each) (a) X k 2 + k + 1 2 k 3- k + 7 (b) X (- 1) k ln k k (c) X 2 k k k (d) X 1 k (ln k ) Solution (a) We use the limit comparison to compare the above series to 1 /k : lim k ( k 2 + k + 1) / (2 k 3- k + 7) 1 /k = lim k k 3 + k 2 + k 2 k 3- k + 7 = 1 2 . Since 0 < 1 / 2 < , it follows that the above series diverges, since 1 /k diverges. (b) Note that lim k ln k k = 0 . Therefore, by the alternating series test, (- 1) k ln k/ k converges. (c) Well use the root test: lim k 2 k k k 1 /k = lim k 2 k = 0 . Since 0 < 1, by the root test, the series converges. The ratio test also works; try it. (d) Well apply the integral test; the series converges if and only if Z 2 1 x ln x dx converges. You can check that this integral diverges (it was one of your homework problems). Therefore, the series diverges as well. Problem 2 (30 points) . Determine the convergence set for the following power series and state their radius of convergence. (10 points each) 1 2 CALCULUS 153: SAMPLE FINAL SOLUTIONS (a) X 2 k x k k ! (b) X (- 1) k kx k k 2 + 1 (c) X 3 k x k Solution (a) lim k 2 k +1 | x | k +1 / ( k + 1)!...
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- Fall '07