We want to find all the values of x
such that the given series would converge
which is called the
interval of convergence
for the power series.
To do this, we'll apply
the ratio test given here
where we know this limit
must be less than one
for the series to converge.
This will give us an open
interval of convergence
and then from there, we'll
test the two end points
to determine if the series
converges or diverges
at the end points.
So notice that a sub n would be equal to
negative one to the power of n
times x to the power of n
divided by eight to the power of n
times the quantity n squared plus nine
and therefore, a sub n plus one
would be equal to negative
one to the power of n plus one
times x to the power of n plus one
divided by eight to the
power of n plus one times
here we'd have the quantity
n plus one squared plus nine.
Let's go ahead and multiply this out.
We would have n squared plus
two n plus one plus nine
or n squared plus two n plus 10.
So the next step, we'll
show it in this form here.
So now we'll take the limit
as n approaches infinity
of the absolute value of
a sub n plus one divided by a sub n
but instead of dividing here,
we'll multiply by the
reciprocal of a sub n.
So first, we have a sub n plus one
and then times the reciprocal of a sub n
so we'd have eight to the nth
times the quantity n
squared plus nine divided by
negative one to the nth x to the nth.
Now, let's simplify.
Notice how we have n plus one
factors of x in the numerator
and only n factors of x in the denominator
so this simplifies to one.
We have one remaining factor
of x in the numerator.
Same thing for the base of negative one.
We have n factors of negative one here
and plus one factors of negative one here.
This simplifies to one.
This simplifies to one
factor of negative one
and then for base eight,
we have n factors of eight here,
n plus one factors of eight here.
This simplifies to one.
This simplifies to one factor of eight.
So now we have the limit
as n approaches infinity
of the absolute value of
this would be negative x
times the quantity n squared plus nine
and the denominator would be eight
times the quantity n squared
plus two n plus 10.
And looking at this limit
here, notice how in terms of n,
we have degree two numerator
and the degree two denominator
and because the degrees are the same
as n approaches infinity,
this limit approaches the
ratio leading coefficients
so this limit is equal to
the absolute value of
negative x over eight.
In order for this series to converge,
this limit must be less than one.
So if we solve this
absolute value inequality,
we can determine the open
interval of convergence.
So we can think of this as negative 1/8 x
and the absolute value of
negative 1/8 is positive 1/8
so we can write this as 1/8
times the absolute value of x
less than one, multiply
both sides by eight.
Absolute of x is less than eight
and therefore, x is less than eight and
also has to be greater than negative eight
which means the open
interval of convergence
would be from negative
eight to positive eight.
Notice how this interval
is centered at zero
which is also where the
infinite series is centered
since we have x here,
not x minus the constant c.
But we still have to test the end points
for convergence or divergence.
Let's do this on the next slide.
So notice when x equals negative eight,
we would have the summation
from n equals one to infinity of
negative one to the nth
times negative eight to the nth
divided by eight to the nth
times the quantity n squared plus nine.
Well, negative one to the nth
times negative eight to the nth.
These would always have the same sign
so we can write this as negative one times
negative eight to the nth
or just eight to the nth.
So we'd have the summation
from n equals one to infinity
of eight to the nth
divided by eight to the nth
times the quantity n squared plus nine
and notice how this simplifies as well.
This simplifies to one.
So now we have the summation
from n equals one to infinity
of one dived by n squared plus nine.
Notice how this does resemble a p-series
where p would be equal to two.
It's not listed here
but if we did the direct comparison test
and compare this to the summation
from n equals one to infinity
of one divided by n squared
which we know converges,
notice how these terms would
have smaller denominators
and therefore, the terms of this series
would be smaller than the
terms of this converging series
and therefore, by the
direct comparison test,
the series converges.
So we'll say since
one divided by the quantity
n squared plus nine
is less than or equal to
one divided by n squared
by the direct comparison test,
the series converges at
x equals negative eight.
And now we'll test that x equals eight
so that would give us the summation
from n equals one to infinity
of negative one to the nth
times eight to the nth
divided by eight to the nth
times n squared plus nine.
Notice how this simplifies to one
so we have the alternating series
of negative one to the nth divided by
the quantity n squared plus nine.
We'll notice how this series
is the absolute value of this series
and therefore if this converges,
this will also converge
or we could apply the
alternating series test
where again, a sub n
would be equal to one divided
by n squared plus nine
which is greater than zero.
The limit as n approaches
infinity of a sub n
does equal zero and a sub n plus one
is less than or equal to sub n
because as n increases,
these terms get smaller and smaller.
So we'll say by the
alternating series test,
the series converges at x equals eight.
So because this series
converges at both end points,
the interval of convergence
is not this open interval.
It would be the closed interval
from negative eight to positive eight.
So the interval of convergence
includes both end points.
So looking at the homework question,
we would say the series is convergent
from x equals negative eight
and it does include the
left end point so we say yes
to x equals positive eight
and it does include the
right end point so we say yes
for the right end point.
I hope you found this helpful.
