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May 23, 2011 S E C T I O N 10.3 Convergence of Series with Positive Terms 661 35. summationdisplay n = 1 n 3 n solution Choose N such that n 2 n for n N . Then summationdisplay n = N n 3 n summationdisplay n = N parenleftbigg 2 3 parenrightbigg n The latter sum is a geometric series with r = 2 3 < 1, so it converges. Thus the series on the left converges as well. Adding back in the finite number of terms for n < N shows that summationdisplay n = 1 n 3 n converges. summationdisplay n = 1 n 5 2 n 37. Show that summationdisplay n = 1 sin 1 n 2 converges. Hint: Use sin x x for x 0. solution For n 1, 0 1 n 2 1 < π ; therefore, sin 1 n 2 > 0 for n 1. Moreover, for n 1, sin 1 n 2 1 n 2 . The series summationdisplay n = 1 1 n 2 is a p -series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude that the series summationdisplay n = 1 sin 1 n 2 also converges. Does summationdisplay n = 2 sin ( 1 /n) ln n converge? In Exercises 39–48, use the Limit Comparison Test to prove convergence or divergence of the infinite series. 39. summationdisplay n = 2 n 2 n 4 1 solution Let a n = n 2 n 4 1 . For large n , n 2 n 4 1 n 2 n 4 = 1 n 2 , so we apply the Limit Comparison Test with b n = 1 n 2 . We find L = lim n →∞ a n b n = lim n →∞ n 2 n 4 1 1 n 2 = lim n →∞ n 4 n 4 1 = 1 . The series summationdisplay n = 1 1 n 2 is a p -series with p = 2 > 1, so it converges; hence, summationdisplay n = 2 1 n 2 also converges. Because L exists, by the Limit Comparison Test we can conclude that the series summationdisplay n = 2 n 2 n 4 1 converges. summationdisplay n = 2 1 n 2 n 41. summationdisplay n = 2 n radicalbig n 3 + 1 solution Let a n = n radicalbig n 3 + 1 . For large n , n radicalbig n 3 + 1 n n 3 = 1 n , so we apply the Limit Comparison test with b n = 1 n . We find L = lim n →∞ a n b n = lim n →∞ n n 3 + 1 1 n = lim n →∞ n 3 radicalbig n 3 + 1 = 1 . The series summationdisplay n = 1 1 n is a p -series with p = 1 2 < 1, so it diverges; hence, summationdisplay n = 2 1 n also diverges. Because L > 0, by the Limit Comparison Test we can conclude that the series summationdisplay n = 2 n radicalbig n 3 + 1 diverges.
May 23, 2011 662 C H A P T E R 10 INFINITE SERIES summationdisplay n = 2 n 3 radicalbig n 7 + 2 n 2 + 1 43. summationdisplay n = 3 3 n + 5 n(n 1 )(n 2 ) solution Let a n = 3 n + 5 n(n 1 )(n 2 ) . For large n , 3 n + 5 n(n 1 )(n 2 ) 3 n n 3 = 3 n 2 , so we apply the Limit Comparison Test with b n = 1 n 2 . We find L = lim n →∞ a n b n = lim n →∞ 3 n + 5 n(n + 1 )(n + 2 ) 1 n 2 = lim n →∞ 3 n 3 + 5 n 2 n(n + 1 )(n + 2 ) = 3 . The series summationdisplay n = 1 1 n 2 is a p -series with p = 2 > 1, so it converges; hence, the series summationdisplay n = 3 1 n 2 also converges. Because L exists, by the Limit Comparison Test we can conclude that the series summationdisplay n = 3 3 n + 5 n(n 1 )(n 2 ) converges.

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