HOMEWORK ASSIGNMENT 1 NOTESDAVID BEN MCREYNOLDS1. Before We S tartWhen we solve problems, one important aspect of problem solvingthat is often overlooked is believing that you can solve the problem.If you do not think that you can do these problems, then you need toconvince yourself otherwise. You must be confident in your ability.2. Section 10.2This section deals primarily with the question of whether or not asequence of real numbers is bounded and/or monotone. Recall that asequence is bounded if you can find real numbers M1and M2suchthatM1≤ an≤ M2A sequence is monotone if is satisfies on the the following:(1)an≤ an+1(2)an+1≤ anIn the first case, we say the se quence anis nondecreasing. If theinequality is strict, we say the sequence is increasing.In the second case, we say the sequence anis nonincreasing. If theinequality is strict, we say the sequence is decreasing.Lastly, we say a sequence anis eventually monotone if there existsN, a positive integer, such that the sequence anis monotone whenn ≥ N.Date: January 20, 2002.12 DAVID BEN MCREYNOLDS2.1. Problems 1-8. These problems are pattern recognition coupledwith testing your ability to construct a formula.Example 2.1. Consider the sequence2, 5, 10, 17, 26, . . .We see thata1= 2a2= 2 + 3a3= 2 + 3 + 5a4= 2 + 3 + 5 + 7a5= 2 + 3 + 5 + 7 + 9......an= 2 + 3 + 5 + ··· + (2n − 1)We will see later that we can write this in a more compact notation,byan= 2 +nXi=2(2i − 1)2.2. Problems 9-40. These problems comprise the bulk of the firstassignment. We work a few examples:Example 2.2. Consider the sequencean=1n!Now, since the denominator is growing and the numerator is fixed, wewould conjecture that the numbers in the sequence would get small.This tells us that the sequence should be bounded. We show this asfollows:1 ≤ n!, if n ≥ 11 ≥1n!, if n ≥ 1So we have shown that the sequence is bounded above by 1. As forbelow, since the sequence is positive, we have0 <1n!So that we have shown the sequence is bounded.HOMEWORK ASSIGNMENT 1 NOTES 3For monotone, we have two possible techniques that we could try.We could take derivatives of the function we get by replacing n withx, or we can look at ratios.Important: Any time you see the factorial, n!, YOU SHOULDALWAYS THINK RATIO!So, we consider the ratioan+1an=1(n+1)!1n!This equals1(n + 1)!·n!1Now, the reason we think of ratios with the factorial is the following:(n + 1)! = (n + 1)n!So, using this, we get1(n + 1)n!·n!1=1n + 1Since n ≥ 1, we know that n + 1 > 1. Thus1n + 1< 1So, our sequence is monotone decreasing.2.3. Problems 41 and 42. Here is a hint to these problems. DoProblem 41 first. Try using the ideas we used in the above exampleinvolving the factorial. Then try doing problem 42 like problem 41.3. Section 10.3You need to read section 10.3 if you have not. Some importantresults are the following:Theorem 3.1. Every convergent sequence is bounded.A very important result follows by taking the contrapositive of theabove theorem.Theorem 3.2. Every seque nce that is unbounded is not convergent.That is, unbounded sequence are divergent.This is quite useful in showing a sequence is not convergent. ThePinching Theorem or Squeeze Theorem is important as well. Infact, you can use this to do almost every problem in 1-36.4 DAVID BEN MCREYNOLDS3.1. Problems 1-36. Try using the Squeeze Theorem and Theorem 3.2.3.2. Problems 51-58. Try rewriting the sequences in a more familiarform.Example 3.1 (Problem 51). We are given that a1= 1 andan+1=1eanWriting out the first few terms,a1= 1a2=1e· 1 =1e=1e2−1a3=1ea2=1e·1e=1e2=1e3−1a4=1ea3=1e·1e2=1e3=1e4−1......an=1en−1So that the sequence is reallyan=1en−1We know that1 ≤ en−1when n ≥ 1. Recall thate ≈ 2.71818 > 1So,1en−1≤ 1Also, we obviously have0 <1en−1So that our sequence is bounded.HOMEWORK ASSIGNMENT 1 NOTES 5To so that the sequence is monotone, consider the ratio of consecutiveterms.an+1an=1en1en−1=1en·en−11=1een−1·en−11=1e< 1So that our sequence is monotone decreasing. Thus, our sequence con-verges.Remark 1. Notice, when we have a sequence of the form1cnwhere c is a real numb e r, that looking at ratios works quite well. Keepthis as a general philosophy for these and later problems.Now, the question wants us to find the limit. If you notice, we did alot of work just to show it was convergent.Moral: Monotone and bounded does not give you the limit.We could have just noticed that the sequence1en−1has constant numerator and growing denominator. Thus, the sequenceconverges to zero. We could also use the squeeze theorem (though Ithink it is a bit artificial), and0 <1en−1≤1nYou would need to verify thaten−1≥ nfor all positive integers n.Also, notice that from the definition of our sequence (at the beginningof the problem), we can see that it is decreasing, sincean+1an=1eanan=1e< 16 DAVID BEN MCREYNOLDS4. Bonus ProblemsI would like the bonus to be turned in before the first exam. Whenall the bonus problems are turned in, I will send out solutions.Bonus. An imp ortant asymptotic relation for the factorial is the fol-lowing, which is a special case of Stirling’s Formula:n! ∼ nne−n√2πnwhere ∼ meanslimn−→∞n!nne−n√2πn= 1Use this to show the sequencean=n!n2nis bounded. Does this sequence converge?Bonus. Here is an easier ques tion. Give an example for each of thefollowing:(a) A sequence of real numbers that is bounded but not convergent.(b) A sequence of real numbers, ansuch thatan= 1an infinite number of times, but the sequence is not bounded.(c) A sequence of real numbers, ansuch that a2nis a convergentsequence but anis not.(d) A sequence of real numbers, ansuch thatbn= an+1− anis a convergent sequence but anis not.(e) Sequences of real numbers anand bnsuch that anand bnareboth convergent sequences butcn=anbnis not.References[Nickalls] Salas, Hille, and Etgen. Calculus: One and Several Variables. Wiley andSons. 1999.University of Texas at Austin, Department of MathematicsE-mail address :