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18 Nov 2019
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Using the divergence test, determine whether the series diverges or that the divergence test is in-conclusive. (a) sigma^infinity_n = 10 sin(pi/n) (b) sigma^infinity_n = 1 n^2/n^2 - 2n + 5 Determine whether the series converges or diverges. (a) 1 + 1/2 squareroot 2 + 1/3 squareroot 3 + 1/4 squareroot 4 +... (b) sigma^infinity_n = 1 1/(2n)^2 (c) sigma^infinity_n = 1 1/4 squareroot 2 (d) sigma^infinity_n =1 2k^-3/2 Use the Comparison Test or Limit Comparison Test to determine if the following series converge or diverge. If it converges, find the limit. (a) sigma^infinity_n = 1 sin^2 n/n^2 (b) sigma^infinity_n = 1 1/2 squareroot 3n^4 + 1 (c) sigma^infinity_n = 1 1/squareroot n^2 + 1 Use the Alternating Series Test to determine if the following series converge or diverge. If it converges, find the limit. (Make sure to conditions). (a) sigma^infinity_n = 1 (-1)^n n/n^2 + 1 (b) sigma^infinigy_n = 1(-1)^n n^2/n^2 + n + 1 Use the Ratio Test to determine if the following series converge or diverge. If it converges. find the imit. (a) sigma^infinity_n = 1 2^n/n! (b) sigma^infinity_n = 1 3^n/n^2 (c) sigma^infinity_n = 1 e^-n(n^3 + 5) Use the Root Test to determine if the series sigma^infinity_n = 1 (-2n)^5n/(n + 1)^5n converge or diverge. If it converges. find the limit. Determine which test would be most appropriate for the following series. Do not test the series for convergence but provide a short justification. (a) sigma^infinity_n = 1 n - 1/n^3 + 1 (b) sigma^infinity_n = 1 (-1)^n n^2 - 1/n^2 + 1 (c) sigma^infinity_n = 1 n^2n/(1 + n)^3n (d) sigma^infinity_n = 1(-1)^n -1 n^4/4^n (e) sigma^infinity_n = 1 |sin(2n)|/|1 + n^2| (f) sigma^infinity_n = 1 (-1)^n + 1/squareroot n - 1
Please help
Using the divergence test, determine whether the series diverges or that the divergence test is in-conclusive. (a) sigma^infinity_n = 10 sin(pi/n) (b) sigma^infinity_n = 1 n^2/n^2 - 2n + 5 Determine whether the series converges or diverges. (a) 1 + 1/2 squareroot 2 + 1/3 squareroot 3 + 1/4 squareroot 4 +... (b) sigma^infinity_n = 1 1/(2n)^2 (c) sigma^infinity_n = 1 1/4 squareroot 2 (d) sigma^infinity_n =1 2k^-3/2 Use the Comparison Test or Limit Comparison Test to determine if the following series converge or diverge. If it converges, find the limit. (a) sigma^infinity_n = 1 sin^2 n/n^2 (b) sigma^infinity_n = 1 1/2 squareroot 3n^4 + 1 (c) sigma^infinity_n = 1 1/squareroot n^2 + 1 Use the Alternating Series Test to determine if the following series converge or diverge. If it converges, find the limit. (Make sure to conditions). (a) sigma^infinity_n = 1 (-1)^n n/n^2 + 1 (b) sigma^infinigy_n = 1(-1)^n n^2/n^2 + n + 1 Use the Ratio Test to determine if the following series converge or diverge. If it converges. find the imit. (a) sigma^infinity_n = 1 2^n/n! (b) sigma^infinity_n = 1 3^n/n^2 (c) sigma^infinity_n = 1 e^-n(n^3 + 5) Use the Root Test to determine if the series sigma^infinity_n = 1 (-2n)^5n/(n + 1)^5n converge or diverge. If it converges. find the limit. Determine which test would be most appropriate for the following series. Do not test the series for convergence but provide a short justification. (a) sigma^infinity_n = 1 n - 1/n^3 + 1 (b) sigma^infinity_n = 1 (-1)^n n^2 - 1/n^2 + 1 (c) sigma^infinity_n = 1 n^2n/(1 + n)^3n (d) sigma^infinity_n = 1(-1)^n -1 n^4/4^n (e) sigma^infinity_n = 1 |sin(2n)|/|1 + n^2| (f) sigma^infinity_n = 1 (-1)^n + 1/squareroot n - 1
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Trinidad TremblayLv2
29 Mar 2019
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