Section 10.4 Reading Assignment: Comparison Tests

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for and what is really about. Please be careful with this assignment.

References: Thomas' Calculus: Early Transcendentals | Calculus | Calculus | Mathematics | Store | Pearson+

Section 7.4 Assignment Reference for Section 10.4 Reading Assignment:

Section 7.4 Reading Assignment: Relative Rates of Growth1 Instructions. Read through this assignment and complete the three exercises below based on this reading. In Chapter 10, we will encounter lots of different limits as n - co. Fortunately for us, limits at infinity will be the only ones we're concerned with in this chapter, but there is a lot going on with these limits and some intuition that it'd be good to develop about these limits. Hopefully you remember L'Hopital's rule and the fact that - is an indeterminate form, meaning that its value depends on the particulars of the problem. This illustrates that the use of the infinity symbol as a single quantity actually removes a lot of information about a limit. To take an example, while both 3n and n' approach co (as n - co), n is, in a sense, faster than 3n. This can be seen when taking the limit of the fraction lim 3n = lim - = 0. This is what I meant before: there are many different speeds that n+00 n2 n+00 n we can approach infinity, so - can have any value (including infinity). In lim = = 0, the expressions 3n and n' are "racing" with each other: 3n in the numerator is pulling the limit closer to infinity, while n in the denominator is pulling the limit closer to zero. The result shows n completely won that race. In other cases, there's something like a tie in the race, which we will explore later. We take this as an inspiration for defining what we mean by one sequence growing faster than another: Definition. A sequence an is growing slower than another sequence by which is written an 0 and b, > 0 for all n 2 N (N an integer). 1. If lim b " = cand c > 0. then Za, and 2b, both converge or both diverge. 2. If lim " = 0 and Eb, converges, then Za, converges. 3. If lim " = oo and 2b, diverges, then Ed,, diverges.Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 57a and h. Since c/2 > 0, there exists an integer / such that Limit definition with whenever n >N. e - c/2, L - e, and a. replaced by a./be Thus, for n > N. 2 2 If Eb,, converges, then E(3c/2)b, converges and Za, converges by the Direct Comparison Test. If 2b,, diverges, then E(c/2)6,, diverges and Za,, diverges by the Direct Comparison Test. EXAMPLE 2 Which of the following series converge, and which diverge? (a) + + 7 + 9 20 + 1 2n + 1 + iME 16 25 - = (n + 1)2 in' + 2n + 1Chapter 10 Infinite Sequences and Series 10.4 Comparison Tests 609 +4 + + ...=27- (b) + 3+ 7+ 15 (e) 1 + 2 In 2 1+313 9 + 1 + 4 In 4 14 21 1 + 5 Solution We apply the Limit Comparison Test to each series. (a) Let a, = (2n + 1)/(wr + 2n + 1). For large n, we expect a, to behave like 27/m = 2 since the leading terms dominate for large n, so we let by = 1. Since Eh, = ME diverges and lim " = lim 2n' + n = 2, a poor + 2n + 1 In, diverges by Part 1 of the Limit Comparison Test. We could just as well have taken be = 2, but 1 is simpler. (b) Let a, = 1/(2" - 1). For large a, we expect a,, to behave like 1/2", so we let b = 1/2". Since 1 Con converges and On = lim 2" lim = lim 1-+03 1 - (1/2") 1,Ia, converges by Part 1 of the Limit Comparison Test. (c) Leta, = (1 + n In n)/(n' + 5). For large n, we expect a, to behave like (n In n)) = (Inn), which is greater than 1 for a 2 3, so we let b, = 1. Since 18 diverges and lim = 00, 7+ 5 Ed, diverges by Part 3 of the Limit Comparison Test. EXAMPLE 3 Does In a converge? Solution Because In a grows more slowly than of for any positive constant c (Section 10.1, Exercise 115), we can compare the series to a convergent p-series. To get the p-series, we see that In n n1/4