Welcome to part two of Power Series.
In this video, we'll take a look at some additional examples
of determining the radius and interval of convergence
when the power series is not centered at x equals zero.
So we take a look at the power series here.
Notice we have x minus two to the power of n.
This tells us the power series is centered
at x equals two.
Because our formula here consists of exponentials,
we'll go ahead and find the radius of convergence
by applying the ratio test.
So we'll have the limit as n approaches infinity
of the absolute value of a sub and plus one
divided by a sub n.
To determine a sub n plus one,
we'll just replace n with n plus one.
So we'll have x minus two to the n plus one power
all over n plus one plus one, that would be n plus two
times three to the n plus one.
Now since this is in fraction form,
instead of dividing by a sub n, we'll multiply it
by the reciprocal of a sub n.
So we'll have n plus one
times three to the nth in the numerator,
and we'll have x minus two to the nth in the denominator.
Now let's go ahead and simplify this.
Notice we have n plus one factors of x minus two
in the numerator,
and n factors of x minus two in the denominator.
That will leave one factor of x minus two
in the numerator.
The factors of n plus one and n plus two don't simplify,
so we'll have n plus one here, n plus two here.
Then looking at the factors of three,
there's n factors of three in the numerator
and n plus one factors of three in the denominator
leaving one factor of three in the denominator.
Now as n approaches infinity, n plus one
over n plus two approaches one,
so we're left with the absolute value of x minus two
divided by three must be less than one
in order for this power series to converge.
Now let's go ahead and factor out the one third here.
So we'll have one third times the absolute value of x
minus two less than one.
Now we'll multiply both sides by three.
So that will give us the absolute value
of x minus two must be less than three.
Remember, this is centered at x equals two,
so this tells us the radius of convergence
is equal to three.
Since this power series is centered at x equals two,
and the radius of convergence is equal to three,
the open interval of convergence would be two minus three
all the way to two plus three which means the open interval
would be from negative one to positive five.
Now we need to test the endpoints to see
if this power series converges at x equals negative one
and x equals five.
Let's do that on the next slide.
Let's first take a look when x is equal to negative one.
So our numerator would be negative three to the nth.
Our denominator would be the same.
Notice we have negative three to the nth in the numerator
and three to the nth in the denominator.
Well this would simplify to negative one to the nth
divided by n plus one.
Now the question becomes does this converge or diverge.
If it converges we can include negative one
in the interval of convergence, and if it doesn't we can't.
This is an alternating series.
Let's go ahead and apply the alternating series test
as we see here.
So we'll take the limit as n approaches infinity
of a sub n, which is one over n plus one.
This does equal zero.
That's the first check.
Now need to make sure that zero is less than
a sub n plus one, less than or equal to a sub n.
Well, a sub n plus one would be one over n plus one
plus one or n plus two.
And a sub n is one over n plus one.
Well, this fraction has larger denominators
than this fraction here.
Therefore, one over n plus two will always be
less than or equal to one over n plus one.
So both conditions are met.
This series does converge when x equals negative one.
We can include the endpoint of negative one
in the interval of convergence.
Now we need to check x equals five.
We'll replace x with five so we'll have five minus two
to the nth, or three to the nth all over n plus one
times three to the nth.
This simplifies nicely.
We have the summation of one over n plus one.
This looks very similar to one over n which does diverge.
But we can't use the direct comparison test
because these terms here are actually going to be less
than one over n.
If we're trying to show that they diverge
by the comparison test, we'd need the terms to be larger.
We could use the limit comparison test,
let's go ahead and use the integral test where f of x
will be equal to one over x plus one.
The anti-derivative of one over x plus one
would be natural log x plus one.
So we'll have the limit as b approaches infinity
of natural log b plus one minus natural log
one plus one or natural log two.
Natural log b plus one approaches infinity,
as b approaches infinity so this limit does not exist.
Therefore, this series diverges at x equals five.
We knew from the radius of convergence
that the open interval of convergence was
from negative one to five, but now we can include
negative one while the interval stays open at five.
Let's go ahead and take a look at this last problem.
To identify where the power series is centered,
it must be in the form of x minus c.
This tells us the power series is centered
at x equals negative one.
Because we have a factorial and an exponential part,
we'll go ahead and apply the ratio test to determine
the radius of convergence.
So we'll have the limit as n approaches infinity
of the absolute value of a sub n plus one.
That will be x plus one to the power of two times n plus one
divided by n plus one factorial.
Now instead of dividing by a sub n,
we're going to multiply by the reciprocal of a sub n.
We'll have n factorial here and then we'll have
x plus one to the power of two n in the denominator.
Let's go ahead and simplify this.
We know n plus one factorial would be n plus one
times n times n minus one and so on.
And n factorial is n times n minus one and so on.
This simplifies nicely to one factor of n plus one
in the denominator.
Now if we look at this power on x plus one,
this will be two n plus two factors of x plus one.
Down here we have two n factors of x plus one.
So we have two extra factors of x plus one
in the numerator.
As n approaches infinity, the denominator increases
without bound, so this limit is equal to zero.
Regardless of the value of x,
this limit will always equal zero,
and therefore the radius of convergence
would be equal to infinity which tells us
that the interval of convergence would be
from negative infinity to positive infinity.
We do have to be careful here that when we have this limit
equal to zero, it does not mean the radius
of convergence is equal to zero.
It means that it converges for all values of x,
and therefore the radius is equal to positive infinity
meaning the interval of convergence would be
from negative infinity to positive infinity.
Hope you found this helpful, thank you for watching.