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### Fall 2001 - Fitzsimmons' Class - Exam 1

Course: MATH 142A, Fall 2001
School: UCSD
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Word Count: 453

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142A (P. Math Fitzsimmons) First Midterm Solutions October 17, 2001 1. (a) A sequence {an } is monotone provided it is increasing (i.e., an+1 an for all n) or decreasing (i.e., an+1 an for all n). (b) A sequence {an } is bounded provided there are numbers L and U such that L an U for all n. Thus, a sequence is bounded if it is both bounded above and bounded below. (c) The completeness property of the real...

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142A (P. Math Fitzsimmons) First Midterm Solutions October 17, 2001 1. (a) A sequence {an } is monotone provided it is increasing (i.e., an+1 an for all n) or decreasing (i.e., an+1 an for all n). (b) A sequence {an } is bounded provided there are numbers L and U such that L an U for all n. Thus, a sequence is bounded if it is both bounded above and bounded below. (c) The completeness property of the real number system is this: If an increasing sequence is bounded above, then it converges. 2. Use the definition of limit (of a sequence) to show that 2n2 = 2. n 2 + n2 lim Solution. We begin by estimating the difference between the nth term of the sequence and its limit: () 2n2 4 4 -4 = -2 = 2. 2 + n2 2 + n2 2 + n2 n Now, given > 0, choose the "cutoff" number n to be larger than 2/ . Thus, if n n then n2 > 4/, so 4/n2 < , and so (by the estimate in ()) 4 2n2 - 2 2 < . 2 2+n n 3. Find the indicated limits using one or more of the Limit Theorems (sum, product, quotient, squeeze, etc.). (a) lim n + sin(n) n n+1 (b) lim 2n n 2n + 3n Solution. (a) Because -1 sin n 1, we have n + sin(n) n-1 1 n+1 n+1 so 1 - n-1 n + sin(n) 1. 1 + n-1 n+1 Now limn n-1 = 0, and limn [1 n-1 ] = 1 by the Sum Theorem, so the lower bound (1 - n-1 )/(1 + n-1 ) converges to 1 by the Quotient Theorem. Therefore, the by Sandwich Theorem, the limit in question is 1. (b) Dividing numerator and denominator by 3n we see that lim 0 (2/3)n 2n = lim = = 0, n n (2/3)n + 1 +3 0+1 n 2n because limn (2/3)n = 0. 1 4. Let {an } be a bounded sequence. Show that n an lim = 0. n Solution. Since {an } is bounded, there are constants A and B such that A an B for all n. This yields the "sandwich" A an B , n n n n 1. Now observe that limn (A/ n) = 0: Given > 0, choose as cutoff n any number larger than (A/)2 . Similarly, limn (B/ n) = 0. It now follows from the Sandwich Theorem that limn (an / n) = 0. 5. Consider the sequence defined recursively by: a1 = 1, an+1 = an /(1 + a2 ), n = 1, 2, . . .. Notice that n an > 0 for each n 1. (a) Show that {an } is a decreasing sequence. (b) Show that L = limn an exists, and find the value of L. Solution. (a) Since an > 0 for all n, an+1 = an an < = an , 1 + a2 1+0 n for all n. Thus, the sequence {an } is therefore (strictly) decreasing. (b) Because {an } is decreasing and bounded below (by 0), it has a limit L 0. Sending n to infinity in the recursion defining {an }, we find that () The inequality 1 <1 1 + L2 if L>0 L= 1 L =L . 1 + L2 1 + L2 in conjunction with () implies that L = 0. Thus, L = 0 is the only solution of (), so limn an = 0. 2
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