- [Phillip] So, we're
gonna find the interval
of convergence for this power series.
And we start the same
way as we would radius
of convergence which is using our,
applying our ratio test
which is basically a sub n plus one,
so we're gonna put in n
plus one for all of our ns
times the reciprocal of a sub n.
And then we simplify and see
what we have to work with.
So, we'll rewrite a
couple of those exponents
so we can see real clearly
our simplifications.
So, x to the n plus n
is x to the n times x.
Three to the n n squared
over this will be three
to the n times three
times n plus one squared x to the n.
Our cancels now are really clear
because we can see that x to the n
will cancel with that one
and this three to the n
will cancel this one and that's
about all we're gonna get.
This no longer depends on n,
so I'm gonna pull it out of the limit
but it does still have an
absolute value around it.
Now what do we have left in our limit?
We have n squared on top
over three times n plus
one squared on bottom.
We have an infinity over infinity case,
so we can use L'Hopital's rule.
You might already know what the limit is
but let's just go through it.
Absolute value of x
means we're gonna take the derivative
of the top and bottom separately.
So, top that gives us two n,
on bottom it gives us six
times n plus one.
Now we still have an
infinity over infinity case,
so we can actually if we want
use L'Hopital's rule again
and we'll have the absolute value of x
and we'll recalculate our limit
using the derivatives of the
numerator and denominator.
Derivative on top is just two,
derivative on bottom is just six.
And what we can see is
this is gonna be equal
to the absolute value of x times 1/3.
We want this to be less
than one, excuse me,
we want this to be less,
getting ahead of myself,
we want this to be less than one
in order for this to converge,
so we'll multiply both sides by three.
That's where the three is gonna come from.
We're gonna get the absolute value of x
must be less than three.
And so, this is actually
our radius of convergence
but in this case, we're
gonna go all the way
to define the interval.
Well, if the absolute value
of x must be less than three,
that means x has to sit in between three
and negative three.
So, we have our endpoints.
The only thing we don't know for sure yet
is if we wanna include out endpoints
because at our endpoints,
the ratio test comes out to be one
which means it's inconclusive,
so we're gonna test
both of these endpoints.
We need to test our series
at x equals negative three
and at x equals three.
So, what we do is we go
back to our series here
and we plug those values in for x.
So, if I plug in the negative three,
I get the summation
of negative three to the n
over three to the n n squared
and then we rewrite it to see
what's going to happen.
So, these guys here, I can write
to the same n power.
And so what I end up with
is negative three over three to the n
and I'm just gonna leave the
one over n squared over here
and that gives me negative one to the n
times one over n squared.
And if you notice, I have
an alternating series
and we can also say that the limit
as n goes to infinity
of one over n squared equals zero,
so this is convergent
by the alternating series test,
so we want to include negative three.
So, I put a equals there.
Now let's see what happens
when we include three.
It's gonna be very similar.
We're gonna get three to the n
over three to the n
and I'm gonna split up the
one over n squared again.
And we can see that this
is actually just going
to be one,
so our series is just one over n squared
but we know that's
convergent by our p-series.
So, we're actually gonna
include that endpoint as well,
so our interval of convergence
is negative three to three
including both endpoints.
