Math 1B: Discussion ExercisesGSI: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Find two or three classmates and a few feet of chalkboard. As a group, try your hand at thefollowing exercises. Be sure to discuss how to solve the exercises — how you get the solution ismuch more important than whether you get the solution. If as a group you agree that you allunderstand a certain type of exercise, move on to later problems. You are not expected to solve allthe exercises: some are very hard.Exercises marked with an § are from Single Variable Calculus: Early Transcendentals for UCBerkeley by James Stewart. Others are my own or are independently marked.Ratio and Root TestsThe Ratio and Root Tests have a very similar form. LetPanbe a series. The Ratio Test beginsby asking you to compute limn→∞an+1an, and the Root Test begins by asking you to computerlimn→∞np|an|. It is a theorem that if both these limits exist, then they are equal (for a givenseries, either one or both may not exist, in which case the corresponding tes t is inconclusive), solet’s call the common limit L. Then if L < 1, the seriesPanconverges absolutely, and if L > 1,then the series diverges. If L = 1, the tests are inconclusive.In general, when L = 1, it is because the seriesP|an| is comparable to a P-seriesP1/np(although this is certainly not always the case). If so, then you can decide whetherP|an| convergesor diverges, and hence whetherPanis absolutely convergent or not. Of course, ifPanis absolutelyconvergent, then you can stop testing; if it is not absolutely convergent, then it may still beconditionally convergent, and a test like the Alternating Series Test may prove thatPanconverges.1. § Determine whether the following series are absolutely convergent, conditionally convergent,or divergent.(a)∞Xn=1n!100n(b)∞Xn=1(−1)n(1.1)nn4(c)∞Xn=1n√n3+ 2(d)∞Xn=1(−1)ne1/nn3(e)∞Xn=1sin 4n4n(f)∞Xn=1n22nn!(g)∞Xn=1(−2)nnn(h)∞Xn=2−2nn + 15n(i)∞Xn=2n(ln n)n2. § Prove thatP∞n=0xn/n! converges for any x. What does this say about limn→∞xn/n!?3. (a) It is a fact that limn→∞1 +1nn= e. Use this fact and the Ratio Test to determine ifthe series∞Xn=1n!nnconverges or diverges.(b) It turns out that for the above series, the limit in the Root Test also exists. What is thelimit in the Root Test, and what does part (a) say about its value?(c) Find all numbers x such that the Ratio and Root Tests are inconclusive when appliedto the series:∞Xn=1xnn!nn1(d) Use part (b) to justify the following estimate:n! ∼nnen(e) The estimate in part (d) is the leading-order part of Stirling’s Formula, which says:n! ≈√2πnnenor, more precisely,limn→∞n!√2πnnen= 1Use this estimate to determine whether the following series converge of diverge:i.∞Xn=1enn!nnii.∞Xn=1nnenn!iii.∞Xn=1(−n)nenn!For part iii., you may assume that {nn/(n!en)} is a monotonic sequence.4. § For which positive integers k does the series∞Xn=1(n!)2(kn)!converge?5. § LetPanbe a series with positive terms, and let rn= an+1/an. Suppose that limn→∞rn=L < 1, so thatPanconverges by the Ratio Test. Let Rnbe the remainder after n terms, i.e.Rn= an+1+ an+2+ an+2+ . . . .(a) If {rn} is a decreasing sequence and rn+1< 1, show, by summing a geometric series,that Rn≤an+11 − rn+1.(b) If {rn} is an increasing sequence, show that Rn≤an+11 − L6. (a) Let’s say that limn→∞|an+1/an| = L > 1. Prove that limn→∞an6= 0, and thus proveone part of the Ratio Test. Hint: Let rn= |an+1/an|; then there is some N so thatfor n ≥ N , rn> (L + 1)/2 = K > 1. Use this fact to prove that for n ≥ N, we have|an| ≥ Kn|aN|. Since K > 1, limn→∞Kn|aN| = ∞, and so lim |an| = ∞.(b) Let’s say that limn→∞np|an| = L > 1. Prove that limn→∞an6= 0, and thus prove onepart of the Root Test.(c) How would you modify these proofs if limn→∞|an+1/an| = limn→∞np|an| = +∞?7. In this exercise, you’ll prove the other half of the Ratio Test.(a) LetPanbe a series with positive terms, and let rn= an+1/an. Suppose that limn→∞rn=L < 1, and let K = (L + 1)/2. Prove that there is some number N such that for n ≥ N,we have rn≤ K.(b) Conclude that for n ≥ N, we have an≤ Kn|aN|.(c) Use the comparison test and a geometric series to prove thatP∞n=Nanconverges. Con-clude thatP∞1anconverges. Hint: K < 1 (you must prove this).(d) Now letPanbe a series with possibly negative terms. Explain why parts (a)–(c) provethat if lim |an+1/an| < 1, thenPanconverges absolutely.8. Modify and repeat the previous exercise to prove the Root
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