We want to find all the values of x
such that the given series would converge,
which is called the
interval of convergence
of the power series.
We'll begin by applying
the ratio test given here
where we know this limit
must be less than one
in order for the series to converge.
This will give us an open
interval of convergence
and then we'll test the endpoints to see
if the series converges or
diverges at the two endpoints.
This information will give us
the interval of convergence.
So, before we apply the ratio test,
notice that a sub n would be equal to
negative one raised to the power of n
times 10 to the nth times x to the nth
divided by
the square root of n plus two.
We are taking the absolute value here
so we could leave off
negative one to the nth
but I'm going to go ahead and show that.
Now for a sub n plus one,
we would have,
negative one to the power of n plus one
times 10 to the power of n plus one,
times x to the power of n plus one,
divided by, we'd have the square root,
of n plus one
plus two.
Now we'll find 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 by a sub n,
we'll multiply by the reciprocal instead.
So first we have a sub n plus one,
times the reciprocal of a sub n.
Now let's begin simplifying.
Looking at the factors of x,
notice how we have n plus one
factors of x in the numerator,
and n factors of x in the denominator.
We have one more factor
of x in the numerator.
So this simplifies to one,
this simplifies to one factor of x.
The same thing happens for base 10,
we have one more factor
of 10 in the numerator.
So this simplifies to one,
this simplifies to 10,
and then for negative one,
negative one to the nth simplifies to one.
We have one remaining factor
of negative one in the numerator.
So then we have the limit,
as n approaches infinity
of the absolute value of,
we'd have negative 10x
times the square root of n plus two.
In the denominator,
we just have the square root of n plus one
plus two.
Loking at this limit as
n approached infinity
notice how we have the
square root of n here
and we have the square
root of n plus one here,
the plus one is irrelevant
as n approaches infinity.
So we can think of the denominator
as having n to the power of 1/2
and think of the numerator as
having n to the 1/2 as well.
So we can think of the
degrees of the numerator
and denominator as being the same
and therefore this limit is equal
to the ratio of the leading coefficients.
Which in this case would
be negative 10x over one,
or just negative 10x.
So this is equal to the
absolute value of negative 10x
which if this is going to converge,
must be less than one.
So if we solve this
absolute value inequality
we can determine the open
interval of convergence.
So we have the absolute
value of negative 10x
must be less than one.
Since the absolute value
of negative 10 is 10,
we can write this as 10 times
the absolute value of x,
less than one dividing by 10,
you have the absolute
value of x less than 1/10.
Which means x is less than 1/10
and
x is greater than negative 1/10.
So the open interval of convergence
is from negative 1/10 to positive 1/10.
But it still may converge
at one or both endpoints.
So now we'll test the endpoints.
So first when x equals negative 1/10,
we would have the summation
from n equals one to infinity
of negative one to the nth
times 10 to the nth times x to the nth
which would be negative 1/10 to the nth
divided by the square root of n plus two.
Let's simplify this.
If we have 10 to the nth,
times negative 1/10 to the nth,
because they're both being
raised to the nth power
we can write this as 10
times negative 1/10 to the nth.
Which would be negative one to the nth.
So this is equal to
negative one to the nth.
So negative one to the nth
times negative one to the nth
would be negative one to the
power of n plus n, or two n.
So we have negative one to the two n
divided by the square root of n plus two.
But notice for all values of n,
we're raising negative
one to an even power
which means this will
always be positive one.
So this simplifies even
more to the summation
from n equals one to infinity
of just one divided by the
square root of n plus two.
Now to determine if this
converges or diverges,
notice how this resembles the series
where we have the summation
from n equals one to infinity
of one divided by the square root of n.
Which we can write as the summation
of one divided n to the 1/2,
which we know diverges
by the P series test,
with P equals 1/2.
So now we'll apply the
limit comparison test
to hopefully show this series
here is also divergent.
To apply the limit comparison test,
we'll now find the limit
as n approaches infinity of a sub n,
which is one divided by the
square root of n plus two.
Instead of dividing by b sub n,
multiply by the reciprocal.
So we'd have times the
square root of n over one.
So we have the limit as
n approaches infinity of,
the square root of n over the
square root of n plus two.
Notice as n approached infinity,
this would approach positive one,
which is positive and finite.
Therefore because this
series was divergent,
our series is also divergent.
So to summarize this we'll say,
by the limit comparison test,
the series
diverges
at x equals negative 1/10.
Now for x equals positive 1/10,
we would have the summation
from n equals one to infinity of,
the same thing we have here
except this would be positive 1/10.
Which means we'd have
negative one to the nth
times positive one to the nth
divided by
the square root of n plus two.
Notice how one to the nth
is always just positive one,
so we have the alternating series
where we have the summation
from n equals one to infinity
of negative one to the nth
divided by the square root of n plus two.
So if we want to apply the
alternating series test,
notice that a sub n is equal to one
divided by the square root of n plus two,
which is always greater than zero.
The limit
as n approaches infinity
of a sub n equals zero
and notice as n increases
these fractions will
get smaller and smaller
because the denominators
would be increasing.
So, a sub n plus one is less
than or equal to a sub n
and therefore by the
alternating series test,
the series converges
at x equals positive 1/10.
So the series diverges at negative 1/10
but converges at positive 1/10.
Therefore the interval of convergence
would remain open on negative 1/10,
but it would be closed at positive 1/10.
So, this is the interval of convergence
and therefore for the homework question
the series is convergent
from x equals negative 1/10,
it does not include the left endpoint
because the interval is
open on negative 1/10,
so this is no,
to x equals positive 1/10.
It includes positive 1/10 so
the right endpoint is included,
so here we have yes.
I hope you found this helpful.
