9:59:59.000,9:59:59.000 1520 11.5 #2 Alternating Series Examples 9:59:59.000,9:59:59.000 https://youtu.be/nSbe28QQSR0 9:59:59.000,9:59:59.000 We're going to try to determine[br]whether this alternating series 9:59:59.000,9:59:59.000 converges or diverges[br]using the alternating series test. 9:59:59.000,9:59:59.000 The first thing we have to figure out [br]is actually a formula for b-n. 9:59:59.000,9:59:59.000 If you recall, b-n is basically the [br]absolute value of each of the terms. 9:59:59.000,9:59:59.000 And if we're looking at this, 9:59:59.000,9:59:59.000 we just ignore the part [br]that causes the sign to alternate, 9:59:59.000,9:59:59.000 Hopefully you would agree[br]that b-n is just 1 over n. 9:59:59.000,9:59:59.000 The first term is 1, 9:59:59.000,9:59:59.000 the second term we're [br]subtracting is one half [1/2], 9:59:59.000,9:59:59.000 and then one third [1/3], 9:59:59.000,9:59:59.000 then we subtract one fourth [1/4], [br]and we add one fifth [1/5]. 9:59:59.000,9:59:59.000 That would be our expression for b-n. 9:59:59.000,9:59:59.000 There are two conditions [br]that need to be met for this test. 9:59:59.000,9:59:59.000 So the first condition is [br]that the n plus first term 9:59:59.000,9:59:59.000 is supposed to be less than [br]or equal to the nth term 9:59:59.000,9:59:59.000 for all values of n beyond a certain point. 9:59:59.000,9:59:59.000 And to test that, we just have to figure [br]out what the n plus first term would be; 9:59:59.000,9:59:59.000 of course, that would be 1 over n plus 1. 9:59:59.000,9:59:59.000 If we compare that to the nth term, [br]which is 1 over n, 9:59:59.000,9:59:59.000 clearly, 1 over n plus 1 [br]is less than 1 over n, 9:59:59.000,9:59:59.000 so that satisfies the first condition. 9:59:59.000,9:59:59.000 The second condition is that the limit as [br]n goes to infinity for b-n needs to equal 0, 9:59:59.000,9:59:59.000 so that's the next thing to test. 9:59:59.000,9:59:59.000 And in some examples, [br]we'll actually do this first 9:59:59.000,9:59:59.000 because if this is not true, [br]then the whole test is going to fail. 9:59:59.000,9:59:59.000 But in this case, if we look at the limit [br]as n goes to infinity for 1 over n, 9:59:59.000,9:59:59.000 hopefully everybody would [br]agree that that definitely is 0. 9:59:59.000,9:59:59.000 Since this is an alternating series [br]and these two conditions have been met, 9:59:59.000,9:59:59.000 that implies that this series right here, 9:59:59.000,9:59:59.000 just like we drew out the diagram [br]of in the first video, converges. 9:59:59.000,9:59:59.000 This series, n goes from 1 to infinity, 9:59:59.000,9:59:59.000 negative 1 to the n minus 1 [br]divided by n converges, 9:59:59.000,9:59:59.000 and it converges by[br]the alternating series test. 9:59:59.000,9:59:59.000 We've got another alternating series here. 9:59:59.000,9:59:59.000 This one starts with a negative term, 9:59:59.000,9:59:59.000 but the formula that we have [br]is a little bit different. 9:59:59.000,9:59:59.000 You can see I've listed out [br]the first few terms. 9:59:59.000,9:59:59.000 I've chosen not to reduce all the fractions 9:59:59.000,9:59:59.000 just so that we can see the pattern [br]that we've got going on here, 9:59:59.000,9:59:59.000 we're going to use the [br]alternating series test 9:59:59.000,9:59:59.000 to try to determine whether [br]this series converges or not. 9:59:59.000,9:59:59.000 To begin with, [br]let's figure out what b-n would be. 9:59:59.000,9:59:59.000 That's the absolute value of each term. 9:59:59.000,9:59:59.000 Basically, the only thing that affects [br]the sign here is this part. 9:59:59.000,9:59:59.000 That means the b-n would just [br]be 3n divided by 4n plus 1. 9:59:59.000,9:59:59.000 Now, it's not immediately obvious 9:59:59.000,9:59:59.000 if these terms are actually [br]shrinking as n goes to infinity, 9:59:59.000,9:59:59.000 we could look at the first few [br]and try to figure out 9:59:59.000,9:59:59.000 whether those fractions [br]are getting smaller or not, 9:59:59.000,9:59:59.000 but I would actually suggest ignoring [br]step number 1 for the time being 9:59:59.000,9:59:59.000 (just because that's a tougher [br]question to answer), 9:59:59.000,9:59:59.000 and let's look at step 2. 9:59:59.000,9:59:59.000 Let's try to figure out if the limit [br]as n goes to infinity for b-n is equal to 0. 9:59:59.000,9:59:59.000 So if we actually write[br]in the formula for b-n, 9:59:59.000,9:59:59.000 we're going to wind up with 3n[br]divided by 4n plus 1. 9:59:59.000,9:59:59.000 And to do this limit, we can just divide [br]everything by the highest power of n, 9:59:59.000,9:59:59.000 which is actually just n to the first. 9:59:59.000,9:59:59.000 Now, of course, the n’s are going [br]to cancel in the first two terms, 9:59:59.000,9:59:59.000 but this last term is going to wind up [br]approaching 0 as n goes to infinity, 9:59:59.000,9:59:59.000 and we're going to be left with [br]3 over 4 plus 0, or three fourths [3/4], 9:59:59.000,9:59:59.000 and that is clearly not equal to 0. 9:59:59.000,9:59:59.000 Since we failed this second condition, 9:59:59.000,9:59:59.000 that actually means that the [br]alternating series test doesn't apply, 9:59:59.000,9:59:59.000 so we may as well not [br]even try to figure out 9:59:59.000,9:59:59.000 whether that first condition is met or not. 9:59:59.000,9:59:59.000 But how do we determine whether [br]the series converges or not? 9:59:59.000,9:59:59.000 Well, fortunately, back in Section 11.2, 9:59:59.000,9:59:59.000 we found out about something [br]called the test for divergence, 9:59:59.000,9:59:59.000 and what that says is, if these terms [br]right here of the original series 9:59:59.000,9:59:59.000 do not approach 0, then that means [br]that the series would be divergent. 9:59:59.000,9:59:59.000 If we were to look at the limit as n [br]goes to infinity of the original terms, 9:59:59.000,9:59:59.000 (negative 1 to the n, [br]times 3n ,divided by 4 plus 1), 9:59:59.000,9:59:59.000 what we would wind up finding out 9:59:59.000,9:59:59.000 is that the absolute value[br]of the terms approach 3/4. 9:59:59.000,9:59:59.000 But because of this [br]alternating portion here, 9:59:59.000,9:59:59.000 that means that for large values of n, 9:59:59.000,9:59:59.000 we're going to be approaching [br]numbers that are close to positive 3/4 9:59:59.000,9:59:59.000 and then negative 3/4, 9:59:59.000,9:59:59.000 and then positive 3/4, [br]and then negative 3/4. 9:59:59.000,9:59:59.000 Since that means the terms are not [br]actually coalescing around a single value, 9:59:59.000,9:59:59.000 what does that tell us about this limit? 9:59:59.000,9:59:59.000 Well, what that tells us [br]is that this limit does not exist. 9:59:59.000,9:59:59.000 And if we look back at [br]the test for divergence, 9:59:59.000,9:59:59.000 it says that if the terms [br]approach any limit other than 0 9:59:59.000,9:59:59.000 or if the limit of the terms does not exist, 9:59:59.000,9:59:59.000 that means that the series [br]is going to be divergent. 9:59:59.000,9:59:59.000 Therefore, by the test for divergence, 9:59:59.000,9:59:59.000 it's actually not the alternating [br]series test that tells us this result; 9:59:59.000,9:59:59.000 it's actually the test for divergence. 9:59:59.000,9:59:59.000 Because of that, we can say [br]that this series has to diverge 9:59:59.000,9:59:59.000 because if the individual [br]terms don't approach 0, 9:59:59.000,9:59:59.000 then the series automatically diverges.[br]