We want to find all the values of x
such that the given series would converge.
This is called the interval of convergence
where the interval of
convergence of a power series
is the interval of x-values
which the series converges.
To find the interval of convergence,
we first apply an infinite
series convergence test.
This gives us an open
interval of convergence
and then we check the endpoints
to determine if the series
converges or diverges.
This information will give us
the interval of convergence.
So we'll apply the ratio test
to determine the open
interval of convergence
where we know this limit
must be less than one
in order for the series to converge
and then we'll test the endpoints
to see if the series
converges or diverges.
So first, notice that a sub n
would be equal to two to
the nth times x to the nth
times the quantity n plus one
divided by the quantity n plus nine
and therefore, a sub n plus one
would be equal to two to
the power of n plus one
times x to the power of n plus one
times the quantity n plus
one plus one or n plus two
divided by the quantity n plus
one plus nine or n plus 10.
So now, we'll find this limit
so we'll have the limit
as n approaches infinity
of the absolute value of a sub n plus one.
Now, instead of dividing by a sub one,
we'll multiply by the reciprocal
so we'd have times the
quantity n plus nine
divided by two to the nth x to the nth
times the quantity n plus one.
Now, looking at the base two,
notice how we have n factors of two here,
n plus one factors of two here.
We have one more factor
of two in the numerator.
So this simplifies to one.
This simplifies to one factor of two.
The same thing for the xs.
We have n factors of x here,
n plus one factors of n here.
We have one more factor
of x in the numerator
so this simplifies to one.
This simplifies to one factor of x.
So now, we have the limit
as n approaches infinity of
the absolute value of two x
times the quantity n plus two
times the quantity n plus nine
divided by the quantity n plus 10
times the quantity n plus one.
Looking at just the n part, notice how
we have a degree two n
polynomial in the numerator
and a degree two n polynomial
in the denominator as well
so the degrees are the same
and therefore, as n approaches infinity,
the n part approaches the ratio
of the leading coefficients
which would be one over one or just one.
So this limit is going to equal
the absolute value of two x
but if we want to, we can show more work
by multiplying the n part out
which would give us the limit
as n approaches infinity of
the absolute value of two x
times the quantity.
We'd have n squared
plus 11 n plus 18
divided by the quantity n squared
plus 11 n
plus 10.
Again, notice how for the n
part, the degrees are the same
so limit is equal to
two x times one over one
or just the absolute value of two x.
In order for this series
to converge though
by the ratio test,
this limit must be less than one.
So if we solve this
absolute value inequality,
we can determine the open
interval of convergence.
So if we factor out two,
we'd have two times
the absolute value of x
is less than one, divide
both sides by two.
Absolute value of x is less than 1/2
which means x is less than 1/2
and it also has to be
greater than negative 1/2
which means the open
interval of convergence
would be the open
interval from negative 1/2
to positive 1/2.
Notice how this is centered at zero.
Notice how the interval of
convergence is centered at zero.
We can tell by divisional series
since we have x here, not
x minus the constant c.
But now, we have to test the endpoints
to see if the series converges or diverges
at negative 1/2 and positive 1/2.
So when x is negative 1/2,
we'd have the summation from
n equals one to infinity of
we'd have two n times negative 1/2
to the n times the quantity n plus one
divided by the quantity n plus nine.
Well, if we have two
raised to the power of n
times negative 1/2
raised to the power of n,
we could write that as two
times negative 1/2 to the nth
which should be negative one to the nth.
So we can write this as the summation
from n equals one to infinity of
negative one to the nth
times the quantity n plus one
divided by the quantity n plus nine.
Notice how we do have an
alternating series where a sub n
would be equal to the quantity n plus one
divided by the quantity n plus nine
but notice how this fails
the nth term divergence test
because the limit as n
approaches infinity of a sub n
doesn't equal zero.
The limit as n approaches infinity of
the quantity n plus one
divided by the quantity n plus nine
would be equal to one
which doesn't equal zero.
Those limits equal to one
because once again, notice how
the degree of the numerator
and denominator are the same.
They're both degree one and therefore,
the limit is equal to the ratio
of the leading coefficients.
One over one or one.
So the series diverges at
x equals negative 1/2
by the nth term divergence test
and x equals positive 1/2.
The same thing happens except now,
we'd have the summation from
n equals one to infinity of
this would now be positive 1/2.
So instead of having
negative one to the nth,
we would have one to the nth
times the quantity n plus one
divided by the quantity n plus nine.
Well, one to the nth is just one
so we have the summation of n plus one
divided by the quantity n plus nine.
Notice that a sub n is the same
and the limit would be one as well
which doesn't equal zero
so the series also
diverges at x equals 1/2
by the nth term divergence test.
And since the series
diverges at both endpoints
then the interval of convergence remains
the open interval from
negative 1/2 to positive 1/2.
So to answer our question, we'll say
the series is convergent
from x equals negative 1/2.
It does not include the left
endpoint so we'll say no.
To x equals positive 1/2
but it does not include
the right endpoint either
so we say no for the right endpoint.
I hope you found this helpful.
