Solutions to Quiz 91. Determine whether the seriesn(ln(n99))101converges.∞Xn=21(A) It converges by the Comparison Test with∞Xn=21n.(B) It diverges by the Integral Test.(C) It diverges by the Comparison Test with∞Xn=21n(ln(n))2.(D) It diverges by the Comparison Test with∞Xn=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,n≥2.The functiony=1991011x(ln(x))101,x≥2,satisfies all assumptions of the Integral Test, and we haveZ∞21991011x(ln(x))101dx=199101Z∞ln(2)1u101du <∞,which implies the convergence of the series.2. The series∞Xn=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 series∑∞n=11√n=∞. We havelimn→∞arctan(n(n+1))√n1√n= limn→∞arctan(n(n+ 1)) =π2= 0.Thus∑∞n=1arctan(n(n+1))√ndiverges.However, the series does converge. To see this, use the Alternating Series Test. Wehavelimn→∞arctan(n(n6