WEBVTT

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1520 11.5 #2 Alternating Series Examples

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https://youtu.be/nSbe28QQSR0

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We're going to try to determine
whether this alternating series

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converges or diverges
using the alternating series test.

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The first thing we have to figure out 
is actually a formula for b-n.

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If you recall, b-n is basically the 
absolute value of each of the terms.

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And if we're looking at this,

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we just ignore the part 
that causes the sign to alternate,

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Hopefully you would agree
that b-n is just 1 over n.

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The first term is 1,

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the second term we're 
subtracting is one half [1/2],

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and then one third [1/3],

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then we subtract one fourth [1/4], 
and we add one fifth [1/5].

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That would be our expression for b-n.

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There are two conditions 
that need to be met for this test.

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So the first condition is 
that the n plus first term

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is supposed to be less than 
or equal to the nth term

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for all values of n beyond a certain point.

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And to test that, we just have to figure 
out what the n plus first term would be;

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of course, that would be 1 over n plus 1.

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If we compare that to the nth term, 
which is 1 over n,

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clearly, 1 over n plus 1 
is less than 1 over n,

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so that satisfies the first condition.

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The second condition is that the limit as 
n goes to infinity for b-n needs to equal 0,

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so that's the next thing to test.

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And in some examples, 
we'll actually do this first

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because if this is not true, 
then the whole test is going to fail.

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But in this case, if we look at the limit 
as n goes to infinity for 1 over n,

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hopefully everybody would 
agree that that definitely is 0.

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Since this is an alternating series 
and these two conditions have been met,

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that implies that this series right here,

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just like we drew out the diagram 
of in the first video, converges.

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This series, n goes from 1 to infinity,

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negative 1 to the n minus 1 
divided by n converges,

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and it converges by
the alternating series test.

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We've got another alternating series here.

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This one starts with a negative term,

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but the formula that we have 
is a little bit different.

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You can see I've listed out 
the first few terms.

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I've chosen not to reduce all the fractions

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just so that we can see the pattern 
that we've got going on here,

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we're going to use the 
alternating series test

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to try to determine whether 
this series converges or not.

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To begin with, 
let's figure out what b-n would be.

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That's the absolute value of each term.

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Basically, the only thing that affects 
the sign here is this part.

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That means the b-n would just 
be 3n divided by 4n plus 1.

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Now, it's not immediately obvious

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if these terms are actually 
shrinking as n goes to infinity,

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we could look at the first few 
and try to figure out

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whether those fractions 
are getting smaller or not,

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but I would actually suggest ignoring 
step number 1 for the time being

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(just because that's a tougher 
question to answer),

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and let's look at step 2.

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Let's try to figure out if the limit 
as n goes to infinity for b-n is equal to 0.

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So if we actually write
in the formula for b-n,

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we're going to wind up with 3n
divided by 4n plus 1.

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And to do this limit, we can just divide 
everything by the highest power of n,

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which is actually just n to the first.

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Now, of course, the n’s are going 
to cancel in the first two terms,

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but this last term is going to wind up 
approaching 0 as n goes to infinity,

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and we're going to be left with 
3 over 4 plus 0, or three fourths [3/4],

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and that is clearly not equal to 0.

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Since we failed this second condition,

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that actually means that the 
alternating series test doesn't apply,

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so we may as well not 
even try to figure out

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whether that first condition is met or not.

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But how do we determine whether 
the series converges or not?

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Well, fortunately, back in Section 11.2,

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we found out about something 
called the test for divergence,

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and what that says is, if these terms 
right here of the original series

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do not approach 0, then that means 
that the series would be divergent.

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If we were to look at the limit as n 
goes to infinity of the original terms,

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(negative 1 to the n, 
times 3n ,divided by 4 plus 1),

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what we would wind up finding out

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is that the absolute value
of the terms approach 3/4.

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But because of this 
alternating portion here,

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that means that for large values of n,

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we're going to be approaching 
numbers that are close to positive 3/4

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and then negative 3/4,

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and then positive 3/4, 
and then negative 3/4.

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Since that means the terms are not 
actually coalescing around a single value,

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what does that tell us about this limit?

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Well, what that tells us 
is that this limit does not exist.

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And if we look back at 
the test for divergence,

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it says that if the terms 
approach any limit other than 0

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or if the limit of the terms does not exist,

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that means that the series 
is going to be divergent.

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Therefore, by the test for divergence,

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it's actually not the alternating 
series test that tells us this result;

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it's actually the test for divergence.

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Because of that, we can say 
that this series has to diverge

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because if the individual 
terms don't approach 0,

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then the series automatically diverges.