Okay, we're going to do an overview
of all of the different convergence tests
that we've learned so far and do
a bunch of examples, one or two
for each kind of test that we've had.
Now, you have to start somewhere
and you have to start by knowing
whether certain things converge.
You have to know geometric series,
you have to know p series.
And you have to be able to do
a little bit of algebra.
So if you've got a quantity like this,
you want to rewrite it as the sum of
three terms, 2^{-n} and 3/n^2 and 5/n^3,
and then you should be able to just
recognize-- oh yeah, that's
a geometric series where the common
ratio is 1/2, so that converges.
3/n^2, that's a p series with p=2,
that converges.
5/n^3, that's a p series with p=3,
that converges.
Convergent plus convergent plus convergent
means the whole thing converges.
What if you had this slightly
different guy?
Well then we would write that as
the sum of 2^{-n} plus 3/n plus 5/n^2.
Geometric converges, p series converges,
p series diverges because this is
a p series with p=1, so this diverges.
So the most basic thing you have
to do is do a little bit of algebra
and compare it to the standard examples,
and of course you have to know
the standard examples.
Now, the first other test we learned
is the divergence test.
What does this series do? Well you
can break it up into pieces
but you should notice that when you
take n very large, this is a tiny number
and this number is very very 
close to three.
If you take n=1000000, you get 3000005
over 1000000, that's close to three.
So if we call this thing a_n, a_n
approaches three, so the sum
of the a_n diverges.
If the things you're adding don't go
to zero, the sum automatically diverges.
So we call that the divergence test.
It's really not a special test, it's
just-- come on, if it doesn't go
to zero, there's no hope of it
converging.
Now let's look at the comparison
and the limit comparison test.
And here is where the geometric and
the p series really shine because
they're the things that you always
compare things to.
If you have the sum of 1/(n^2+n),
you say, 'oh, 1/(n^2+n) &lt; 1/n^2.'
I mean, the numerator is bigger
so the ratio is smaller,
and this is a p series. This is a p series
with p=2, so this converges,
so this converges.
So this converges by comparison
to 1/n^2.
So you often get really complicated
stuff and you shouldn't let
the extra complication phase you.
You say, 'oh, this is behaving like 1/n^2.'
In fact it's less than 1/n^2, so we
do better than 1/n^2,
which means we win.
What about this example?
(n+cos(n))/(n^3+n^2).
This is starting to look a little
bit ugly, you've got this
extra cos(n) upstairs, you've got
this extra n^2 downstairs.
So here's where you use the limit
comparison test.
Let's call this a_n, and we can
compare to b_n, which is 1/n^2.
And let's take the ratio. You take
a_n divided by b_n and
that's going to be (n+cos(n))/(n^3+n^2)
divided by 1/n^2, so that means
times n^2, so that's
(n^3+n^2cos(n))/(n^3+n),
and that's (1+cos(n))/n divided
by 1+(1/n), and that approaches one.
The ratio approaches one, so since
the ratio approaches one,
by the limit comparison test,
either both series converge or
both series diverge.
So the sum of 1/n^2 converges, it's
a p series, this also converges.
So this converges by the limit
comparison test.
Okay, we'll do one more and
then take a break.
The integral test.
If you have a function that's decreasing,
you compare it to an integral.
In this case, you want the integral
of 1/(xln(x)) and you ask
does that converge or not?
You do the u substitution u=ln(x),
du=1/x dx and you get that
this is the integral of 1/u du which
is ln(u) plus a constant,
which is ln(ln(x)) plus a constant
and that diverges.
If you take x big enough, ln(x) will
be as big as you'd like,
so the natural log of that will be
as big as you'd like, this diverges.
Since the integral diverges,
the sum diverges.
It's not a p series, it's not an
exponential so you can't do
a direct comparison, but you do
an integral and you see that it diverges.
Okay, one more.
This you compare it to the integral
of 1/x(ln(x))^2.
You use the exact same substitution
u=ln(x), du=1/x dx, and then
you get that this is the integral of
1/u^2 du, which is -1/u plus a constant.
That's no problem, as x goes to
infinity-- as x goes to infinity,
-1/ln(x) goes to zero, there's no
problem with this improper integral.
The integral converges,
so the sum converges.
We'll do the remainder of our tests
in the next video.
