[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,1520 11.5 #2 Alternating Series Examples
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,https://youtu.be/nSbe28QQSR0
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,We're going to try to determine\Nwhether this alternating series
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,converges or diverges\Nusing the alternating series test.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,The first thing we have to figure out \Nis actually a formula for b-n.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,If you recall, b-n is basically the \Nabsolute value of each of the terms.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,And if we're looking at this,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,we just ignore the part \Nthat causes the sign to alternate,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Hopefully you would agree\Nthat b-n is just 1 over n.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,The first term is 1,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,the second term we're \Nsubtracting is one half [1/2],
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and then one third [1/3],
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,then we subtract one fourth [1/4], \Nand we add one fifth [1/5].
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,That would be our expression for b-n.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,There are two conditions \Nthat need to be met for this test.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,So the first condition is \Nthat the n plus first term
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,is supposed to be less than \Nor equal to the nth term
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,for all values of n beyond a certain point.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,And to test that, we just have to figure \Nout what the n plus first term would be;
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,of course, that would be 1 over n plus 1.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,If we compare that to the nth term, \Nwhich is 1 over n,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,clearly, 1 over n plus 1 \Nis less than 1 over n,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,so that satisfies the first condition.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,The second condition is that the limit as \Nn goes to infinity for b-n needs to equal 0,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,so that's the next thing to test.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,And in some examples, \Nwe'll actually do this first
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,because if this is not true, \Nthen the whole test is going to fail.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,But in this case, if we look at the limit \Nas n goes to infinity for 1 over n,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,hopefully everybody would \Nagree that that definitely is 0.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Since this is an alternating series \Nand these two conditions have been met,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,that implies that this series right here,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,just like we drew out the diagram \Nof in the first video, converges.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,This series, n goes from 1 to infinity,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,negative 1 to the n minus 1 \Ndivided by n converges,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and it converges by\Nthe alternating series test.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,We've got another alternating series here.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,This one starts with a negative term,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,but the formula that we have \Nis a little bit different.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,You can see I've listed out \Nthe first few terms.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,I've chosen not to reduce all the fractions
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,just so that we can see the pattern \Nthat we've got going on here,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,we're going to use the \Nalternating series test
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,to try to determine whether \Nthis series converges or not.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,To begin with, \Nlet's figure out what b-n would be.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,That's the absolute value of each term.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Basically, the only thing that affects \Nthe sign here is this part.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,That means the b-n would just \Nbe 3n divided by 4n plus 1.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Now, it's not immediately obvious
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,if these terms are actually \Nshrinking as n goes to infinity,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,we could look at the first few \Nand try to figure out
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,whether those fractions \Nare getting smaller or not,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,but I would actually suggest ignoring \Nstep number 1 for the time being
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,(just because that's a tougher \Nquestion to answer),
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and let's look at step 2.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Let's try to figure out if the limit \Nas n goes to infinity for b-n is equal to 0.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,So if we actually write\Nin the formula for b-n,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,we're going to wind up with 3n\Ndivided by 4n plus 1.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,And to do this limit, we can just divide \Neverything by the highest power of n,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,which is actually just n to the first.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Now, of course, the n’s are going \Nto cancel in the first two terms,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,but this last term is going to wind up \Napproaching 0 as n goes to infinity,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and we're going to be left with \N3 over 4 plus 0, or three fourths [3/4],
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and that is clearly not equal to 0.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Since we failed this second condition,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,that actually means that the \Nalternating series test doesn't apply,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,so we may as well not \Neven try to figure out
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,whether that first condition is met or not.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,But how do we determine whether \Nthe series converges or not?
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Well, fortunately, back in Section 11.2,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,we found out about something \Ncalled the test for divergence,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and what that says is, if these terms \Nright here of the original series
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,do not approach 0, then that means \Nthat the series would be divergent.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,If we were to look at the limit as n \Ngoes to infinity of the original terms,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,(negative 1 to the n, \Ntimes 3n ,divided by 4 plus 1),
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,what we would wind up finding out
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,is that the absolute value\Nof the terms approach 3/4.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,But because of this \Nalternating portion here,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,that means that for large values of n,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,we're going to be approaching \Nnumbers that are close to positive 3/4
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and then negative 3/4,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,and then positive 3/4, \Nand then negative 3/4.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Since that means the terms are not \Nactually coalescing around a single value,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,what does that tell us about this limit?
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Well, what that tells us \Nis that this limit does not exist.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,And if we look back at \Nthe test for divergence,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,it says that if the terms \Napproach any limit other than 0
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,or if the limit of the terms does not exist,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,that means that the series \Nis going to be divergent.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Therefore, by the test for divergence,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,it's actually not the alternating \Nseries test that tells us this result;
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,it's actually the test for divergence.
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,Because of that, we can say \Nthat this series has to diverge
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,because if the individual \Nterms don't approach 0,
Dialogue: 0,9:59:59.99,9:59:59.99,Default,,0000,0000,0000,,then the series automatically diverges.\N