21c_ps3_soln - Co-21C Problem Solving#3T(Solutions Converge...

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Co-21C Problem Solving #3T (Solutions) Converge or diverge? State which test you use, and be sure all of the prerequisites are satisfied, if applicable. 1) 4 3 n + 2 n = 1 ! " = 4 3 n ! 3 2 n = 1 " # = 4 9 1 3 $ % & ( ) n n = 1 " # , so this is a geometric series with r = 1 3 and r < 1 so it converges . 2) n 3 n 3 + 2 n ! 6 n = 1 " # Since lim n !" n 3 n 3 + 2 n # 6 = 1 $ 0 , the series diverges by the n th term test. 3) n n 3 + 2 n ! 6 n = 1 " # n n 3 + 2 n ! 6 < 1 n 2 " n 3 < n 3 + 2 n ! 6 " n > 3 , so n n 3 + 2 n ! 6 < 1 n 2 n = 3 " # n = 3 " # , which converges ( p -series, p = 2 > 1), so our series also converges by the Comparison Test. Note that even though the original series did not start at n = 3, it will not affect our answer. Also note that we could have used the Limit Comparison Test to get the same result. 4) n 3 + 1 4 n + 5 n = 3 ! " This series “behaves” like n 1/ 3 n = 1 n 2/ 3 n = 1 ! " n = 1 ! " , which diverges ( p -series, p = 2/3 < 1). Use the Limit Comparison Test: lim n !" n 1/ 3 + 1 4 n + 5 1 n 2/ 3 = lim n !" n 1/ 3 + 1 4 n + 5 # n 2/ 3 1 = lim n !" n + n 2/ 3 4 n + 5 = 1 4 ! 0 , so our series diverges .

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