# For all n 2 1 n 2 8 1 n 2 and the series 1 n 2

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6. For alln>2,1n2-8<1n2, and the series1n2converges,so by the Comparison Test, the series1n2-8converges.Correct Answers:CIICII4
18.(1 pt) For each sequenceanfind a number r such thatanrnhas a finite non-zero limit.(This is of use, because by the limit comparison test the seriesn=1anandn=1rnboth converge or both diverge.)n=17n+5(-1)n
B.an=32n3n+nr =C.an=4n+n3+4188n+3n+4r =(1 pt)(a)Carefully determine the convergence of theseriesn=1(-1)n3n. The series is-1)n3n. The series isA. absolutely convergentB. conditionally convergentC. divergentSOLUTIONNote that for both of these series the terms alternate in sign,and the magnitudes of successive terms decrease, so that we areable to apply the alternating series test.
19.(1 pt) Select the FIRST correct reason why the givenseries diverges.-1)n1A. Diverges because the terms don’t have limit zeroB. Divergent geometric seriesC. Divergent p seriesD. Integral testE. Comparison with a divergent p seriesF. Diverges by limit comparison testG. Cannot apply any test done so far in class1.n=1(-1)n(2n)!(n!)22.n=1cos(nπ)ln(6)3.n=11n4.n=1ln(n)n
6.n=17n+5(-1)n
20.(1 pt)(a)Carefully determine the convergence of theseriesn=1(-1)n3n. The series is
(b)Carefullydeterminetheconvergenceoftheseriesn=1(-1)n3n. The series isA. absolutely convergentB. conditionally convergentC. divergentSOLUTIONNote that for both of these series the terms alternate in sign,and the magnitudes of successive terms decrease, so that we areable to apply the alternating series test.
(-1)n1
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