Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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3:21 RealAnalysis-ISBN-fix... was a way to be rigorous without having to run back to the definition each time. Fortunately, there is a way. If we can use the definition to prove some general rules about limits then we could use these rules whenever they applied and be assured that everything was still rigorous. A number of these should look familiar from calculus. Problem 64. Let (c) - (c, c, e,...) be a constant sequence. Show that lim e=e. In proving the familiar limit theorems, the following will prove to be a very useful tool. Lemma 1. (a) Triangle Inequality Let a and b be real numbers. Then la +6 s lal+l6|- (b) Reverse Triangle Inequality Let a and b be real numbers. Then Ja| - 16 S la - M CONVERGENCE OF SEQUENCES AND SERIES 80 Problem 65. (a) Prove Lemma 0/ Hint: For the Reverse Triangle Inequality, consider Jal - la - b+ (b) Show |la| - |6|| s la - 4. / Hint: You want to show lal - 16| s la - b| and /'l9 - | 5 (l9l – l|)- Theorem 7. If lim a, =a and lim b, - 6, then lim (a, +6) = a+b. A We will often informally state this theorem as "the limit of a sum is the sum of the limits." However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We'll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof. SCRAPWORK: If we let e > 0, then we want N so that if n > N, then |(a, + 6) - (a + b)<e. We know that lim n =a and lim, - 6, so we can make Jan - al and b, - 6| as small as we wish, provided we make n large enough. Let's go back to what we want, to see if we can ckose the gap between what we know and what we want. We have |(a, + b) – (a + 6) |-|(a, - a) + (b, - b) |s la, - a|+|b, - b| by the triangle inequality. To make this whole thing less than e, it makes sense to make each part less than 5. Fortunately, we can do that as the definitions of lim, d = a and lim,x b. = b allow us to make a, - a and b. - 6 arbitrarily small. Specifically, since lim,x n = a, there exists an Ni stuch that if n> N, then Ja, -al <. Also since lim, b, = 6, there exists an Ng such that if n> Na then b, - 6 <5. Since we want both of these to occur, it makes sense to let N max(N1, Na). This should be the N that we seek. Problem 66. Prove Theorem 2 Theorem 8. If lim a, =a and lim , - 6, then lim (a,b,) = ab. SCRAPWORK: Given e>0, we want N so that if n> N, then Ja,b-ab < e. One of the standard tricks in analysis is to "uncancel." In this case we will subtract and add a convenient term. Normally these would "cancel out," which is why we say that we will uncancel to put them back in. You already saw an example of this in proving the Reverse Triangle Inequality (Problem 6). In the Next 1 Dashboard Calendar To Do Notifications Inbox
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Transcribed Image Text:3:21 RealAnalysis-ISBN-fix... was a way to be rigorous without having to run back to the definition each time. Fortunately, there is a way. If we can use the definition to prove some general rules about limits then we could use these rules whenever they applied and be assured that everything was still rigorous. A number of these should look familiar from calculus. Problem 64. Let (c) - (c, c, e,...) be a constant sequence. Show that lim e=e. In proving the familiar limit theorems, the following will prove to be a very useful tool. Lemma 1. (a) Triangle Inequality Let a and b be real numbers. Then la +6 s lal+l6|- (b) Reverse Triangle Inequality Let a and b be real numbers. Then Ja| - 16 S la - M CONVERGENCE OF SEQUENCES AND SERIES 80 Problem 65. (a) Prove Lemma 0/ Hint: For the Reverse Triangle Inequality, consider Jal - la - b+ (b) Show |la| - |6|| s la - 4. / Hint: You want to show lal - 16| s la - b| and /'l9 - | 5 (l9l – l|)- Theorem 7. If lim a, =a and lim b, - 6, then lim (a, +6) = a+b. A We will often informally state this theorem as "the limit of a sum is the sum of the limits." However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We'll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof. SCRAPWORK: If we let e > 0, then we want N so that if n > N, then |(a, + 6) - (a + b)<e. We know that lim n =a and lim, - 6, so we can make Jan - al and b, - 6| as small as we wish, provided we make n large enough. Let's go back to what we want, to see if we can ckose the gap between what we know and what we want. We have |(a, + b) – (a + 6) |-|(a, - a) + (b, - b) |s la, - a|+|b, - b| by the triangle inequality. To make this whole thing less than e, it makes sense to make each part less than 5. Fortunately, we can do that as the definitions of lim, d = a and lim,x b. = b allow us to make a, - a and b. - 6 arbitrarily small. Specifically, since lim,x n = a, there exists an Ni stuch that if n> N, then Ja, -al <. Also since lim, b, = 6, there exists an Ng such that if n> Na then b, - 6 <5. Since we want both of these to occur, it makes sense to let N max(N1, Na). This should be the N that we seek. Problem 66. Prove Theorem 2 Theorem 8. If lim a, =a and lim , - 6, then lim (a,b,) = ab. SCRAPWORK: Given e>0, we want N so that if n> N, then Ja,b-ab < e. One of the standard tricks in analysis is to "uncancel." In this case we will subtract and add a convenient term. Normally these would "cancel out," which is why we say that we will uncancel to put them back in. You already saw an example of this in proving the Reverse Triangle Inequality (Problem 6). In the Next 1 Dashboard Calendar To Do Notifications Inbox
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To prove our required result, we used the epsilon-delta definition of convergent.

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