S4 - 92.305 Homework 4 Solutions October 9 2009 Exercise 2.4.1 Complete the proof of Theorem 2.4.6 by showing that if the series 2n b2n diverges then so

# S4 - 92.305 Homework 4 Solutions October 9 2009 Exercise...

This preview shows page 1 - 3 out of 5 pages.

92.305 Homework 4SolutionsOctober 9, 2009Exercise 2.4.1.Complete the proof of Theorem 2.4.6 by showing that if the seriesn=02nb2ndiverges, then so doesn=1bn. Example 2.4.5 may be a useful reference.Proof:We will show that ifn=02nb2ndiverges thenn=1bndoes too by exploiting arelationship between the partial sumssm=b1+b2+· · ·+bmandtk=b1+ 2b2+· · ·+ 2kbk.Becausen=02nb2ndiverges, its monotone sequence of partial sums (tk) must be unbounded.To show that (sm) is unbounded it is enough to show that for allkN, there is termsmsatisfyingsmtk/2. This argument is similar to the one for the forward direction, only toget the inequality to go the other way we group the terms insmso that the last (and hencesmallest) term in each group is of the formb2k.Given an arbitraryk, we focus our attention ons2kand observe thats2k=b1+b2+ (b3+b4) + (b5+b6+b7+b8) +· · ·+ (b2k-1+1+· · ·+b2k)b1+b2+ (b4+b4) + (b8+b8+b8+b8) +· · ·+ (b2k+· · ·+b2k)=b1+b2+ 2b4+ 4b8+· · ·+ 2k-1b2k=12(2b1+ 2b2+ 4b4+ 8b8+· · ·+ 2kb2k)=b1/2 +tk/2.Because (tk) is unbounded, the sequence (sm) must also be unbounded and cannot converge.Therefore,n=1bndiverges.squaresolidExercise 2.4.2.(a) Prove that the sequence defined byx1= 3 andxn+1=14-xnconverges.Proof:We will show that this sequence is decreasing and bounded (which will also settlethe issue of well-definedness: sincex1is 3 and subsequent terms are even smaller,xnneverequals 4, so the denominator 4xnnever vanishes).First let’s use induction to show that this sequence is decreasing. Observe thatx1= 3>1 =x2. We need to prove that ifxn> xn+1thenxn+1> xn+2. Note thatxn> xn+1impliesthatxn<xn+1. Adding 4 to both sides of the inequality gives 4xn<4xn+1. Itfollows that14xn>14xn+1,1 which is precisely what we need to concludexn+1> xn+2.Thus by induction, (xn) isdecreasing.The argument above shows that (xn) is bounded above by 3, so now we’ll use this fact inproving by induction that (xn) is bounded below, and more specifically, thatxn>0 for alln.Clearlyx1>0. Now assumexn>0. Because (xn) is decreasing, we know thatxnx1= 3,which implies thatxn+1=14-xnis positive. By induction, (xn) is bounded below by 0 for allpositive integersN.Therefore this sequence converges by the Monotone Convergence Theorem.squaresolid(b) Now that we know limxnexists, explain why limxn+1must also exist and equal the samevalue.  #### You've reached the end of your free preview.

Want to read all 5 pages?

• Spring '10
• MetCalfe
• Mathematical analysis, Limit of a sequence, Xn
• • • 