Hi all, let's take a look at section 10.4
which primarily deals with alternating
series.
Anytime you have an alternating series,
you're gonna have that -1 to the n power.
That's going to cause terms of your series
to alternate in sign from negative to
positive, to negative to positive.
Now, the first concept that we're
gonna be talking about in this section
is the concept of absolute convergence.
And all that that is saying is that
your series would converge even if it
didn't alternate in sign.
So sometimes the series will converge
simply because it's alternating in sign
from negative to positive
to negative to positive and
those terms end up
cancelling each other out.
But the concept of absolute convergence
is it's saying that a series is
going to be absolutely convergent if the
absolute value of that series converges.
Now the absolute value that would just
get rid of any alternating sign so
you don't care if you have a -1
to the n or -1 to the n + 1.
If you have an absolute value around it,
every term is gonna be positive anyway.
So it's saying that the series would
absolutely converge if clearly
the absolute value of the series
general term converges.
Now conditional convergence is going to
say that a series a sub n converges but
the series have the absolute
value of a sub n diverges.
So the only reason it's converging is
because the term is alternating in sign.
So now I can say, well,
all right, in this case,
a sub n would be conditionally
convergent if the series converges,
but the absolute value
of the series diverges.
Now, in all of our all alternating
series test, will either use
what's called the alternating
series test or the Leibniz's test.
Different books call it different things,
but either way, it's a very simple test.
It's just saying that the alternating
series you have -1 to the n
power times a sub n that that series
is going to converge if the limit as n
approaches infinity of the general term.
And notice that general term is
not including the -1 to the n.
So if the limit as n approaches
infinity of that general term is 0,
then you can say, well,
then the whole series will converge.
Now, when we look at determining
convergence of an alternating
series such as my first problem here.
I have -1 to the n- 1 power,
as n goes from 1 to infinity of 1
over n to the two-thirds power.
Now, since it's an alternating series,
that's my very first thing here is
I noticed the -1 to the n- 1 power.
So I can go in and say, well, okay,
then the alternating series test tells me,
I can take the limit as n approaches
infinity of the non-alternating part.
So I'm taking the limit as n
approaches infinity of 1 over n
to the two-thirds power.
And I can see as n goes to infinity,
1 over infinity to the two-thirds
power is gonna be 1 over infinity
which has a limiting value of 0.
So therefore I can say,
this series -1 to the n- 1 times 1
over n to the two-thirds power
it's a convergent series.
That series converges.
Now you'll see I put, however, so
just because the series converges
doesn't mean you can't think about
more with these alternating series.
I was thinking hypothetically,
what if that series
didn't alternate in sign?
So if I think about the absolute value of
that series, which I talked about up here.
While the absolute value of that series
would do away with that -1 to the n-
1 power.
And so I would say without that it's just
the series 1 over n to the two-thirds.
And instantly you know hey,
that's a divergent p-series,
because the denominator is not
getting large enough, fast enough for
this function to converge to any value.
So you can say, well,
that's a p-series of p equals two-thirds.
And remember, the decision for
convergence in the p-series,
anytime p is a number less than or
equal to 1, it diverges.
If p would have been greater
than 1 it would converge.
So this is a divergent p-series,
the p equals two-thirds.
Now, what does all this mean?
Well, I showed in the first step
that the series converges due to
the alternating series test.
However, if you're looking at
the absolute value of the series,
it's a divergent p-series.
Well, this is the whole concept of
what I've talked about up here,
about a series being
conditionally convergent.
We say for this number 1,
therefore this series -1 to the n-1
times 1 over n to the two-thirds
converges conditionally.
The words converges conditionally
all that means is yes,
your alternating series does converge.
But the only reason it converges is
because of the alternating sign.
If that alternating sign was not there,
it would be a divergent series.
So you say, well, okay,
it's not a divergent series though,
because the alternating sign is there.
Therefore, we say this series
conditionally converges or
converges conditionally.
Now a second problem I
wanna look at with you.
What if I asked you to
determine the convergence of,
again, it's an alternating series
from n equals 1 to infinity -1 to
the n times n to the fourth
over n cubed + 1.
Well again, alternating series, so
I'll use that alternating series test.
Where I say okay, it will converge
if the limit as n approaches
infinity of a sub n is equal to 0.
But notice in this problem even
if you ignore the alternating
part you still have n to
the fourth over n cubed + 1.
You know as n goes to infinity,
this fraction is going to go to infinity
because you have a larger
power of n in the numerator.
It's gonna be infinity over 1,
it's gonna go to infinity.
So any limiting value that's not
0 is going to show that this
alternating series diverges.
So in order to be convergent,
that limit would have to be 0.
Well, clearly,
infinity is certainly not 0.
So we can say since the limit
as n approaches infinity of
the general term of our series is not 0.
That is conclusive evidence to
show that our series diverges.
And you could either say this is
by the alternating series test, or
like I haven't read here.
You could also just say, well, this
diverges due to the test for divergence or
the divergence test.
The divergence test if you remember all it
says is if you try to plug in the limit as
n approaches infinity of a sub n, if it's
not equal to 0, then the series diverges.
Now, if it is equal to 0,
it doesn't say it converges.
The alternating series test would
tell you it does converge if
the limit as x approaches
infinity of a sub n equals 0.
But either by the alternating
series test or the divergence test,
clearly I can say it's not 0,
therefore this diverges.
Now, one other problem in this section and
you'll notice this section is short.
There's only eight problems in
the homework assignment for this section.
My next one I wanna look at here, it's
technically not an alternating series.
But it kind of behaves like
an alternating series.
If I asked you to determine
the convergence of the series from
1 to infinity of the sine of pi
n over 4 divided by n squared.
So think about this with
me in this numerator.
If n is 1, you'd have the sine of pi over
4, which is square root of 2 over 2.
And this 2 would be the pi over 2 be 1.
So this function is gonna go from a
numerator if square root of 2 over 2 to 1,
back to the square root 2 over 2.
Then you need to go down to 0, then you go
to the negative square root of 2 over 2.
When n is 5 at n equals 6,
does it go straight down to -1?
So all you're doing is you're going around
your unit circle in increments of pi over
4 and telling me what the sign is.
So the numerator, it could be 0,
square root of 2 over 2, 1,
back down to 0, back down to
the negative square root of 2 over 2.
The lowest it could ever be is -1.
Now that lowest it could ever be is -1,
the highest it could ever be is 1.
But it could be a certainly a bunch of
other numbers other than that as well 0
and square to 2 over 2,
a negative square root of 2 over 2.
So that kind of leads it
to a problematic point.
It's not quite an alternating series,
but it's similar to one.
So here's what I'm going to do.
I'm going to consider the absolute value
of the general term of this series.
So let's consider that, well,
the general term would be
the absolute value of this sine of
pi n over 4 divided by n squared.
Now what I know is that
the maximum value the sine could
ever take on is a maximum of 1.
Sometimes it's gonna be
the square root of 2 over 2.
Sometimes it's going to be 0, but
the maximum it could ever be is 1.
So I know even the absolute value of
this general term, it is strictly,
and I say strictly but here I have less
than or equal to 1 over n squared.
This is a true statement.
It would also be true to say it's
strictly less than 1 over n squared.
Either way is fine.
Here I have less than or
equal to 1 over n squared.
Now, what do you already
know about 1 over n squared?
Well, we know that the series 1 over
n squared is a convergent p-series.
Remember anytime you have n to
the denominator raised to a power it is
a p-series.
And that test for convergence there is
anytime p is a number greater than 1,
then that denominator is large
enough to make that fraction small.
So that whenever you're trying
to sum up from 1 to infinity,
you're gonna get
a convergent finite answer.
In this case, we say this is a convergent
p-series because p is equal to 2,
which is greater than 1.
Now, since that series converges, and
I've already shown that the absolute value
of the general term of the series I'm
interested in has to be less than that.
So I can say, since the absolute
value of the sine of pi over
4 over n squared is less than or
equal to 1 over n squared.
That's evidence that the series,
the absolute
value of the sine of pi over 4 over n
squared converges by the comparison test.
Remember the comparison
test if you're less than
the general term of
a series that converges.
Then that's evidence that that series
also converges by the comparison test.
Now back to the original series.
If the absolute value of
that series converges,
that is clear evidence that
the original series converges.
And we would say that this shows
that the series of sine of pi over 4
over n squared from n equals 1 to
infinity converges absolutely.
So you don't even need to show
this individual function.
If you can show the absolute value
converges you've already showed that this
converges.
And we say it converges absolutely,
because even if the signs didn't
alternate, it still converges.
So, when the signs alternate,
it definitely converges or
we just say converges absolutely,
pretty cool problem.
Let me know if you have any questions with
any of the homework assignments in this
section.
They're pretty straightforward
once you get used to it, but
it does take some getting accustomed to.
I'll be glad to help you out with that.
