We want to find all the values of x
such that the given series would converge.
This is called the interval of convergence
for the power series.
Looking at our infinite series,
notice how we have the
quantity x minus six here,
which means the interval of
convergence will be centered
at positive six.
We'll first apply the ratio
test given here below,
where we know this limit
must be less than one
for the series to converge.
So this will give us an open
interval of convergence,
and then we'll test the
endpoints to determine
if this series converges or
diverges at the two endpoints.
This will give us the
interval of convergence.
So a sub n is equal to
the quantity x minus six
raised to the nth divided by n
times negative three raised to the nth,
and therefore, a sub n
plus one would be equal to
the quantity x minus six
to the n plus one power
divided by, here we'd have
the quantity n plus one
times negative three raised
to the power of n plus one.
And therefore, for the ratio test,
we would have 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 we first have a sub n plus one.
So we have the quantity of x.
And then we'd have times
the reciprocal of a sub n,
which would be n times
negative three to the nth
divided by the quantity
x minus six to the nth.
Now let's begin simplifying.
Notice how we have
n plus one factors of x
minus six in the numerator
and only n factors of x
minus six in the denominator.
So this would simplify to one,
this would simplify to
one factor of x minus six.
Notice for negative three,
we have n plus one factors
of negative three in the denominator
and only n factors of
three in the numerator.
So this simplifies to one,
this simplifies to one
factor of negative three.
So now we have the limit
as n approaches infinity
of the absolute value.
We have the quantity x minus six times n
divided by in our denominator we have
the quantity n plus one
times negative three.
Notice both n terms are degree one.
And then for the limit
as n approaches infinity
would be equal to the ratio
of leading coefficients
times the remaining factors,
which would give us the
absolute value of x minus six
divided by negative three.
And we know for this series to converge,
this limit must be less than one.
So if we solve this absolute
value inequality for x,
we can determine the open
interval of convergence.
Let's go ahead and rewrite this as
the absolute value of negative 1/3
times the quantity x
minus six less than one.
And since the absolute
value of negative 1/3
is positive 1/3, we can write this as 1/3
times the absolute value of
x minus six less than one,
multiplying both sides by three,
we have the absolute value of x minus six
is less than positive three.
Again, notice how this interval
will be centered at six
and this three here on the right tells us
the radius of convergence would be three.
But to solve this we would have
x minus six less than three
and x minus six greater
than negative three.
So solving here we have
x is less than nine and
x is, adding six, greater
than positive three.
So the open interval
of convergence would be
from three to nine.
But now it can still
converge at the endpoints.
So now we'll test the endpoints.
So when x equals positive three,
notice how here we'd have three minus six,
so we'd have negative three to the nth.
So we have the summation
from n equals one to infinity
of negative three to the nth divided by
n times negative three to the nth.
We'll notice this simplifies to one.
So we have the summation
from n equals one to infinity
of one divided by n.
We probably recognize
this as a harmonic series
which diverges or by the p-series test
with p equals one diverges.
So we'll say by the p-series test,
with p equals one,
the series diverges at x equals three.
And now we'll test the
series of x equals nine.
Notice how when x equals nine,
here we'd have three to
the nth in the numerator.
So we have the summation
from n equals one to infinity
of three to the n,
divided by n times
negative three to the n.
This will simplify nicely.
We can write this as the
summation of three to the n
divided by n.
And then for negative three to the n,
we can write that as negative one to the n
times three to the n.
So notice how here we have three to the n
over three to the n
which simplifies to one.
Now this factor of negative one can be
in the numerator or denominator.
Let's go and move it up to the numerator
and write this as the summation
from n equals one to infinity
of negative one to the nth
divided by n.
So we have an alternating series.
Notice that a sub n
equals one divided by n
which is greater than one.
The limit as n approaches
infinity of a sub n
equals zero and a sub n plus one
is less than or equal to a sub n.
Notice as it increases, the
terms get smaller and smaller,
and therefore they're decreasing,
satisfying the condition
that a sub n plus one
is less than or equal to a sub n.
So by the alternating series test,
the series converges at x equals nine.
So the series was divergent
at x equals three,
but convergent at x equals nine,
and therefore, our interval of convergence
would be open on three but closed on nine.
This is the interval of convergence,
and therefore, we'll say
this series is convergent
from x equals three.
And we don't include the left endpoints
so the answer is no.
Two x equals nine, where we
do include the endpoints,
so here we say yes to the right endpoint.
I hope you found this helpful.
