Quiz 9 Solution - Solutions to Quiz 9 1 Determine whether the series n=2 1 n ln(n99 101 converges(A It converges by the Comparison Test with n=2 1 n(B

# Quiz 9 Solution - Solutions to Quiz 9 1 Determine whether...

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Solutions to Quiz 9 1. Determine whether the seriesn(ln(n99))101converges. X n =2 1 (A) It converges by the Comparison Test withXn=21n.(B) It diverges by the Integral Test.(C) It diverges by the Comparison Test withXn=21n(ln(n))2.(D) It diverges by the Comparison Test withXn=21nln(n).(E) It converges by the Integral Test.Solution: It converges by the Integral Test (the correct answer is (E)). Note that1n(ln(n99))101=1991011n(ln(n))101,n2.The functiony=1991011x(ln(x))101,x2,satisfies all assumptions of the Integral Test, and we haveZ21991011x(ln(x))101dx=199101Zln(2)1u101du <,which implies the convergence of the series.2. The seriesXn=1arctan((-1)nn(n+ 1))n(A) converges absolutely.(B) diverges, because the terms alternate.(C) diverges, because limn→∞arctan((-1)nn(n+ 1))n6= 0.(D) diverges, even thougharctan((-1)nn(n+1))n= 0.(E) converges conditionally, but not absolutely.
Solution: The arctan function is an odd function, i.e., we have arctan(-x) =-arctan(x)for allx. Consequently,Xn=1arctan((-1)nn(n+ 1))n=Xn=1(-1)narctan(n(n+ 1))n,which shows that the series is alternating.The series does not converge absolutely. To see this, use the Limit Comparison Testand compare to the divergent comparison seriesn=11n=. We havelimn→∞arctan(n(n+1))n1n= limn→∞arctan(n(n+ 1)) =π2= 0.Thusn=1arctan(n(n+1))ndiverges.However, the series does converge. To see this, use the Alternating Series Test. Wehavelimn→∞arctan(n(n 6