x x 3 2 1 x 2 as x By the p test we know that 1 1 x 2 dx converges so by the

# Thomas' Calculus: Early Transcendentals

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x/ ( x 3 +2) 1 /x 2 as x → ∞ . By the p -test we know that 1 1 x 2 dx converges, so by the limit comparison test we know that 1 x x 3 + 2 dx also converges and by the comparison test we conclude that the original integral also converges. 3. (25 points) For each of the series below determine whether it converges or diverges. Justify your answers. (a) n =1 n 2 + 5 n ( n + 1)( n + 2)( n + 3) This series diverges. For rational functions the highest power dominates as we go to infinity. So the n th term is asymptotic to n 2 /n 3 = 1 /n as n goes to infinity. Since 1 1 n diverges by the p-test with p = 1, we conclude that the original series diverges by the limit comparison test. (b) n =2 1 n ln n In this case we use the integral test. Observe that dx x ln x = ln(ln x )+ C and lim t →∞ ln(ln t ) = since lim t →∞ ln t = . Therefore the improper integral 2 dx x ln x diverges. By the integral test the series diverges as well.
(c) n =1 n 2 3 n In this case the ratio or root test works well. For example, with the ratio test we have lim n →∞ ( n + 1) 2 3 n +1 · 3 n n 2 = lim n →∞ n + 1 n 2 1 3 = 1 3 Since this ratio is less than 1, the ratio test says that the series converges.