Math 131AReal AnalysisQuiz 5Instructions: You have 30 minutes to complete the quiz. There are two problems, worth atotal of 24 points. You may not use any books or notes. Partial credit will be given for progresstoward correct proofs. Write your solutions in the space below the questions. If you need morespace use the back of the page. Do not forget to write your name and UID in the space below.Name:Student ID number:Question Points Score1 82 83 8Total: 24Problem 1. 8pts.Suppose f : R → R is differentiable and that f0: R → R is bounded.Prove that f is uniformly continuous.Solution: Let > 0.Since f0is bounded there exists an M > 0 such that∀x ∈ R, |f0(x)| ≤ M.Let δ =M, suppose x, y ∈ R and that |x − y| < δ.If x = y we have |f(x) − f(y)| = 0 < . Otherwise, the MVT gives a z between xand y such that f0(z) =f(x)−f(y)x−yand we see that|f(x) − f(y)| = |f0(z)||x − y| ≤ M|x − y| < Mδ = .Problem 2.(a) [4pts.] SupposeP∞k=1akis a series.Define the sequence of partial sums (sn)∞n=1ofP∞k=1ak.(b) [4pts.] SupposeP∞k=1akis a series with sequence of partial sums given by1 −12n∞n=1.Does the seriesP∞k=1akconverge? Why?Solution:(a) sn=Pnk=1ak.(b) Yes, because the sequence of partial sums converges.Problem 3. 8pts.Suppose thatP∞k=1akandP∞k=1bkare two series consisting of positive non-zero terms.Suppose that limn→∞anbn= 0 and thatP∞k=1bkis convergent.Prove thatP∞k=1akis convergent.Help! You should use the comparison test.Solution: Since limn→∞anbn= 0, we can find an N ∈ N so that∀n ∈ N, n > N =⇒anbn< 1,i.e, ∀n ∈ N, n > N =⇒ |an| < |bn|.SinceP∞k=1bkconverges,P∞k=N +1bkconverges.By the comparison theoremP∞k=N +1akconverges, and
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