# Compare with r c dn evaluate your integral with

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Compare with R c dn = (Evaluate your integral with bottom limit c = 1 .) Solution: SOLUTION B. Again, we note that a n = n + 3 1
n = 1 n + 3 n 2 + 6 n + 8 with R c n + 3 n 2 + 6 n + 8 dn . Taking c = 1, we have Z 1 n + 3 n 2 + 6 n + 8 dn = lim n 1 2 ln ( b 2 + 6 b + 8 ) - 1 2 ln ( 15 ) . Because lim n 2 + 6 b + 8 ) diverges, we know that the integral, and thus the sum, diverge. 4. (1 point)
For each of the following series, indicate whether the integral test can be used to determine its convergence or not, and if not, why. A. n = 1 ln ( n )+ 7 n n Can the integral test be used to test convergence? C. no, because the series is not a geometric series
D. no, because the terms in the series are not recursively defined E. no, because the terms in the series are not defined for all n F. yes B. n = 1 cos ( n π ) n 8 Can the integral test be used to test convergence? C. no, because the series is not a geometric series
D. no, because the terms in the series are not recursively defined