1520 11 5 2 Alternating Series Examples Source

1520 11.5 #2 Alternating Series Examples

https://youtu.be/nSbe28QQSR0

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

converges or diverges
using the alternating series test.

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

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

And if we're looking at this,

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

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

The first term is 1,

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

and then one third [1/3],

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

That would be our expression for b-n.

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

So the first condition is
that the n plus first term

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

for all values of n beyond a certain point.

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

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

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

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

so that satisfies the first condition.

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

so that's the next thing to test.

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

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

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

hopefully everybody would
agree that that definitely is 0.

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

that implies that this series right here,

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

This series, n goes from 1 to infinity,

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

and it converges by
the alternating series test.

We've got another alternating series here.

This one starts with a negative term,

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

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

I've chosen not to reduce all the fractions

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

we're going to use the
alternating series test

to try to determine whether
this series converges or not.

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

That's the absolute value of each term.

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

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

Now, it's not immediately obvious

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

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

whether those fractions
are getting smaller or not,

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

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

and let's look at step 2.

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

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

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

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

which is actually just n to the first.

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

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

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

and that is clearly not equal to 0.

Since we failed this second condition,

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

so we may as well not
even try to figure out

whether that first condition is met or not.

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

Well, fortunately, back in Section 11.2,

we found out about something
called the test for divergence,

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

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

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

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

what we would wind up finding out

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

But because of this
alternating portion here,

that means that for large values of n,

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

and then negative 3/4,

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

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

what does that tell us about this limit?

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

And if we look back at
the test for divergence,

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

or if the limit of the terms does not exist,

that means that the series
is going to be divergent.

Therefore, by the test for divergence,

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

it's actually the test for divergence.

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

because if the individual
terms don't approach 0,

then the series automatically diverges.