Alternating series test.
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Alternating series and it turns out that
there's
a particularly nice criterion for when
they converge.
So here's what normally goes by the name
the alternating series test so the
setup of this is as follows I'm going to
suppose that I've got a sequence.
The terms are decreasing and all of the
terms are positive.
And the limit of the nth term as n goes to
infinity is 0.
So these are the assumptions.
And then in that case, I may conclude that
this alternating series that I build from
the sequence a sub n by multiplying by
minus 1 to the n plus 1 power.
This alternating series converges.
Alternating series form a class of series
for
which the limit test is practically the
whole story.
So given these conditions, if the limit of
a sub n is 0, then this
series converges.
But we already know by just the limit test
that if the limit
of the nth term of this series, remember,
is non-zero, then the series diverges.
But to say that this limit is non-zero is
exactly the same
thing as saying that this limit either
doesn't exist or is non-zero.
So, alternating series are a situation
where
just determining the limit of the nth
term,
or at least the absolute value of the nth
term, tells the whole story.
If that limit is 0, then the series
converges,
and if that limit's non-zero, then the
series diverges.
Now, you already know that the if the
limit
of the nth term is not 0, the series
diverges.
So, in this case, what do I have to show?
So this is the statement that I want to
prove.
I start with some sequence which is
decreasing.
All the terms are positive.
The limit of that sequence is 0.
I want to
conclude that this series, this
alternating series, converges.
Well what does that even mean?
What does it mean to show that this series
converges?
Right, I just have to remember the
definition
of this is really something about the
limit.
All right, it's to say that the limit of
this partial sum, the sum k goes from 1 to
n of minus 1 to the k plus 1 times a sub
k, this sequence of partial sums should
converge and I'll usually denote that n
partial sum by just, s sub n.
Well let's look more carefully at these
partial sums.
So let's look at the first few terms of
the sequence of partial sums, right?
What happens when I just plug in a few
values of n here?
What do I get?
Well, when I plug in n equals 1?
It's not too interesting, right?
It's just the sum of the first term, which
is just a
sub 1.
When I plug in n equals 2, that's a little
bit more interesting, right?
That's the sum of the first two terms,
which is a sub 1 minus a sub 2.
Why is the minus in there?
When I plug in k equals 2 here, I get
negative 1 to the 3rd times a sub 2.
Negative 1 to the 3rd is minus 1.
So that's introducing a negative sign in
front of the a sub 2.
So this is a sub 1 minus a sub 2.
What's the third partial sum?
Well the third partial sum is a sub 1
minus a sub 2 plus the next term in
the series which is minus 1 to the 4th
times a sub 3, it's plus a sub 3.
What's the fourth partial sum?
Well that's a sub 1 minus a sub 2 plus a
sub 3 minus a sub 4.
I mean, it's maybe we're not getting a lot
of intuition here.
The fifth one is a sub 1 minus a sub 2
plus a sub 3 minus a sub 4 plus a sub 5.
It's a little
bit hard maybe to tell what's going on.
So, instead of just looking at these
numbers, I could try to look at these
partial
sums on a number line.
And the key fact to remember here is that
these a sub n's are decreasing, so in
something like the fifth partial
sum, a sub 2 is smaller than a sub 1, a
sub 3 is smaller than both of
these still.
Right?
That's going to help us to draw a nice
picture of
what these partial sums look like on the
number line.
So the first partial sum, you know?
Who knows where it is?
It's just a sub 1 and it's somewhere on
the number line.
But then to get from that first partial
sum to the next partial sum to get from s
of 1 over to s of 2, all right?
I just subtract a sub 2.
So here I am subtracting a sub 2, and that
lands me over here
at the second partial sum, right, because
s sub 2 is a sub 1 minus a sub 2.
Well, how do I get to the third partial
sum?
Where is the third partial sum?
Well, to get to s sub 2 to s sub 3, I just
add a sub 3.
And the key fact here is that a sub n is
decreasing
so the amount that I'm moving to the right
with a sub 3 is
less than the amount that I was moving to
the left with a sub 2.
So maybe here is s sub 3, the third
partial sum.
And then to get from s sub 3 to s sub 4, I
have to subtract a sub 4,
but again, the sequence a sub n is
decreasing, so a sub 4 is smaller than
a sub 3, so I'm moving less to the left
than I was
to the right to go from s sub 2 to s sub
3.
So now I subtract a sub 4, and I land here
at s4, the fourth partial sum.
And to get to s sub 4 to s sub 5, I'd be
adding
a sub 5, but a sub 5 is even smaller than
a sub 4.
So maybe that's somewhere over here.
What do I notice?
Well, what I'm noticing here is that the
even partial
sums, s sub 2 and s sub 4 are increasing.
And the odd partial sums, s sub 1, s sub
3, s sub 5, they're decreasing.
I can be a little bit more precise, right?
This sequence, the sequence s sub 2 n
minus 1 is decreasing.
This sequence, s sub 2 n, is increasing.
So these two sequences are monotone.
The s sub evens
is increasing, but bounded above by s sub
1.
And the s of odds are decreasing, but
bounded below by, say, s sub 2.
So I've got monotone and bounded
sequences.
So by the monotone convergence theorem,
the sequence of even
partial sums is increasing and bounded
above, and therefore the
limit exists.
And by the monotone conversion theorem the
sequence of odd partial sums is decreasing
but bounded below, and consequently the
limit of s sub odd exists.
Well, that's awesome, or not.
What did I actually want to prove?
Well, I actually wanted that the limit of
just the partial sums
exists, I mean, I don't, I don't really
care about the sequence
of even partial sums and the sequence of
odd partial sums.
I just want to know what the sequence of
partial sums, does it have a limit?
So how do I know that the odd and the even
sequences converge to the same thing?
So what I know is that the limit of the
even
partial sums exist and the limit of the
odd partial sums exists.
But I just want to know about the limit of
the partial sums all together, right?
The even terms are getting close to
something, the odd
terms are getting close to something but
are all the terms getting close to
something.
Well to analyze this, I'm going to look
instead at
this limit, I'm going to look at the limit
of
the difference of the even and the odd
partial
sums and there's two different ways to
calculate this limit.
On the one hand
[LAUGH]
well I can just calculate what this term
is, this is the sum of the
first 2 end terms, this is the sum of the
first 2n minus 1 terms.
So if I add up the first 2n terms
and subtract all but the last term that's
exactly
the same then as just the 2 nth term,
which in this case is negative a sub 2n.
So this limit is the same as this limit,
but what is
this limit, right?
Well I actually know what that one's equal
to because
I assume that the limit of the nth term is
0.
So
[LAUGH]
my, my assumption right the limit of
negative a sub 2n is
the same as negative the limit of a sub n
which is 0.
But now there's another way that I could
calculate this limit.
This is the limit of a difference, which
is
the difference of the limits provided the
limits exist.
And in this case, they do, right?
I'm assuming that these two limits exist.
So this is the limit of a difference.
This is a difference of the limits.
But now,
[LAUGH].
I just calculated this a moment ago, to be
equal to 0, right?
What I'm saying here is that the limit of
this
sequence, and the limit of this sequence,
differ by 0.
Which is to say that this limit equals
this limit, right?
Which is just to say that the limit of s
sub n exists, because
the limit of the even terms is equal to
the limit of the odd terms.
So all of the terms together are getting
close
to something in particular and that's what
I wanted.
Right the limit of a partial sums exist
and that's
exactly what it means to say that this
series converges
right to say that a series converges is
just to
say that the limit of the sequence of
partial sums exists.
But wait, there's more.
Let's think back to this picture, right?
This diagram
of the number line with the partial sums
drawn on it.
Well, the odd partial sums are decreasing,
and their limit is some value L.
The even partial sums are increasing, and
their limit's the
same thing, it's also L, because the
difference between even and
odd partial sums are controlled by just
the terms in
the sequence a sub k, and those terms have
limit 0.
So the even partial sums are
increasing to L, the odd partial sums are
decreasing to L.
And what this means is that the value of
this series is between these two, right.
Here's the value of the of the series.
Here's the series, and all of the even
partial sums are below it
and all of the odd partial sums are above
this true value, L.
This is
one of the best reasons to care about
alternating series.
So not only is it really easy to
determine if an alternating series
converges, the whole
story just boils down to whether the limit
of a sub n is 0 or not.
But more than that, I can now give
explicit
error bounds on the value of an
alternating series.
If I compute the 2 nth and the 2n minus
1th partial sum, I know the true value
of my series lands between these two
values.
And this makes alternating series great
for machines.
When you're calculating a partial sum,
you're getting error bounds for free.
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