M 408D SERIES PROBLEMSDAVID BEN MCREYNOLDSProblem 1 (Easy-Medium). Determine whether the following seriesconverge or diverge.∞Xn=1n2+ 33n∞Xn=1nn(2n + 1)!∞Xn=1(π − e)nn!nn∞Xn=11n2− 1∞Xn=1sin n5n∞Xn=13n+17n∞Xn=11sin1n∞Xn=2n2+ 1n2ln n∞Xn=1π + n5(3n + 1)!∞Xn=1sin1n∞Xn=1(n!)2(2n)!∞Xn=11n√3∞Xn=1e−n∞Xn=1tan1n∞Xn=13n(n + 3)Problem 2 (Easy-Medium). Evaluate the following limits.limx−→π−2cos xln(sin x)limx−→01ex− 1−1xlimx−→∞rk2+ 1k2xlimx−→∞101xlimx−→06x− 3x4xlimx−→0ax− bxxlimx−→0sin xx1x2limx−→∞x − sin xxlimx−→∞ex+ e2x1xDate: February 6, 2002.12 DAVID BEN MCREYNOLDSProblem 3 (Medium). Determine whether the following series con-verge or diverge. Some of these are ridiculously hard!∞Xn=21(ln n)ln n∞Xn=12n + 1n2+ n∞Xn=1(n!)nnn!∞Xn=1(nn)!nnn∞Xn=122n4n+1∞Xn=112ln n∞Xn=1cos(2nπ)10n∞Xn=1n3+ 3n2+ 3n + 1(n + 1)6∞Xn=13nn!n2nProblem 4 (Hard). Suppose that we have the following for functionsf(x) and g(x):f(0) = 1 g(0) = ∞f0(0)f(0)= 1−g0(0)(g(0))2= 1Then findlimx−→0f(x)g(x)Problem 5 (Easy). Determine whether the following series convergeor diverge.∞Xn=2n ln n + 1n3ln n∞Xn=11 + 2n3n∞Xn=1n!nn∞Xn=11 + n3πn∞Xn=16n+ 14n+10∞Xn=1nn1192n∞Xn=1n + 32n + n2∞Xn=1ln nn2∞Xn=11n1.001Problem 6 (Medium). Suppose that anis a sequence of numbers andlimn−→∞nan= Lwhere 0 < L < ∞, then what can you say about∞Xn=1anM 408D SERIES PROBLEMS 3Problem 7 (Easy). Determine whether the following series convergeor diverge.∞Xn=2(−1)nn ln n∞Xn=12n4n2− 1∞Xn=1(−1)n24n(2n + 1)!∞Xn=1n7nn!∞Xn=1n23n∞Xn=1n32n∞Xn=1(−1)nlnn + 2n∞Xn=11033√n∞Xn=1n√n − 1nProblem 8 (Easy). Determine whether the following series convergeor diverge.∞Xn=1ne−n∞Xn=11nπ∞Xn=1ne−n2∞Xn=11 +1nn∞Xn=11n√n2− 1∞Xn=1n + 1010n + 1∞Xn=11n(n + 3)∞Xn=11n(n + 1)∞Xn=12n+ 12n+1Problem 9 (Easy-Medium). Determine whether the following seriesconverge or diverge.∞Xn=21n ln n ln(ln n)4 DAVID BEN MCREYNOLDSProblem 10 (Easy-Medium). Determine whether the following seriesconverges conditionally, absolutely, or diverges.∞Xn=1(−1)n(n2+ 1)4n4+ 1∞Xn=1(−1)nn∞Xn=1(−1)n(n + 2)3n2+ 5∞Xn=13 + 4(−1)nn3+ 8∞Xn=2(−1)nn ln n∞Xn=2(−1)nln nn∞Xn=1(−1)n(2n + 3)32n(n + 1)!∞Xn=1(−1)ne−3n∞Xn=1(−1)n3√n∞Xn=1(−1)nn√n∞Xn=1cos(nπ)n + 1∞Xn=1(−1)n3nn2n∞Xn=1(−1)ncos(nπ)∞Xn=1(−1)nsin(nπ)∞Xn=1(−1)nlnn + 2nProblem 11 (Limit Comparison). Use the Limit Comparison Test todetermine the convergence or divergence of the series.∞Xn=1nn2+ 1∞Xn=21√n2− 1∞Xn=12n2− 13n5+ 2n + 1∞Xn=1n + 3n(n + 2)∞Xn=11n +√n2+ 1∞Xn=112n− 5∞Xn=11n! + 6n3∞Xn=1n2n3+ 1∞Xn=1n35n4+ 3Problem 12 (Alternating Series Test). Use the Alternating Series testto determine the convergence or divergence of the series.∞Xn=1(−1)n+1n∞Xn=2(−1)nln n∞Xn=1(−1)nn2n2+ 1∞Xn=1(−1)n(2n)!∞Xn=1(−1)n+1√nn + 2∞Xn=11nsin(2n − 1)π2∞Xn=12(−1)n+1en− e−n∞Xn=12(−1)n+1en+ e−n∞Xn=1(−1)n+1ln(n + 1)n + 1M 408D SERIES PROBLEMS 5Problem 13 (Ratio Test). Use the Ratio Test to test for convergenceor divergence of the series.∞Xn=1n!3n∞Xn=1(n!)2(3n)!∞Xn=14nn!∞Xn=1n22n∞Xn=1(2n)!n5∞Xn=13n(n + 1)nProblem 14 (Root Test). Use the Root Test to test for convergenceor divergence of the series.∞Xn=1n2n + 1n∞Xn=12n√n + 1n∞Xn=1n√n − 1n∞Xn=1ln nnn∞Xn=12nn + 1n∞Xn=21ln nnProblem 15 (Integral Test). Use the Integral Test to test for conver-gence or divergence of the given series.∞Xn=11n + 1∞Xn=1e−n∞Xn=1ne−n∞Xn=1e−ncos n∞Xn=1nk−1nk+ c∞Xn=1nn2+ 36 DAVID BEN MCREYNOLDSProblem 16. Do the following series converge or diverge? Why? Showall work!.∞Xn=1nn3+ 2n∞Xn=1n!n2n∞Xn=1n2− nn4+ 1∞Xn=11n! +√n∞Xn=1nn(n + 1)n∞Xn=1√n(n − 1)n∞Xn=21√n ln n∞Xn=1n −√nn3+√n∞Xn=1nn2+1n∞Xn=2n(ln n)n∞Xn=1(2n + 1)2n(n2+ 1)n∞Xn=1e1nn∞Xn=11√n3+ n∞Xn=1(−1)n√nn + 1∞Xn=11n + ln n∞Xn=1n!n2nProblem 17. Compute the f ollowing limits. Show work.limn−→∞1n1nlimx−→0+xln xlimn−→∞1 −1n2nlimn−→∞n312nlimn−→∞n1√nlimx−→0+ln xxlimn−→∞1√nn2limn−→∞1 +1n + 1nlimn−→∞2n1nUniversity of Texas at Austin, Department of MathematicsE-mail address :
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