We want to find all the values
of X such that the given
series would converge.
This is called the interval of convergence
of the power series, or
the interval of convergence
of a power series is
the interval of x-values
for which this series converges.
To determine the interval of convergence,
step one, we'll apply an
infinite series convergence test.
This will give us an open
interval of convergence.
Step two, we'll check the two endpoints
to determine if the series
converges or diverges.
This information will give us
the interval of convergence.
We have a power series in this form here.
It's centered at zero.
If we have a power series in
this form, it's centered at C.
And a power series will
always converge at its center.
For our example, since
we just have just X here,
it's centered at zero, and
now we'll apply the ratio test
to determine the open
interval of convergence.
Looking at our series,
notice that ace of N
would be equal to five
X, raise the power of N,
divided by N to the eighth,
and ace of N plus one
would be equal to five X
to the power of N plus one,
divided by the quantity
N plus one to the eighth.
So now we'll apply the ratio test.
We'll have the limit as
N approaches infinity
of the absolute value
of ace of N plus one,
divided by ace of N, but
instead of dividing by
ace of N, we'll multiply
the reciprocal instead.
So ace of N plus one would be five X,
raise the power of N plus one,
divided by the quantity
N plus one to the eighth
and then times the reciprocal of ace of N,
which would be N to the eighth divided by
five X to the Nth.
Now I'd like to the
get the base of five X.
Notice how we have N plus
one, factors of five X
in the numerator and
only N factors of five X
in the denominator, so notice how we have
one more factor of five
X in the numerator.
This would simplify to one,
and this would simplify to
just one factor of five X.
So now we have the limit
as N approaches infinity
of the absolute value of five X
times N to the eighth, and
looking at the denominator,
notice how if we were to
start multiplying this out,
the first term would be N to the eighth,
plus several more terms,
but when finding this limit,
as N approaches infinity,
we only care about
the highest degree term for N.
Notice how the degrees are the same
and therefore, the N part
of this would approach
the ratio of the leading coefficients,
which would be one over one.
Notice five X is not affected by N,
which means its limit is
equal to the absolute value
of five X, and this will converge
when this is less than one.
So the solution to this
absolute value inequality
will give us the open
interval of convergence
and then we would test the end points.
So to solve this, we
would factor out five.
That would give us five
times the absolute value
of X less than one,
divide both sides by five,
and we have the absolute value
of X is less than one-fifth.
In this form, one-fifth is called
the readies of convergence,
and if we continue to solving,
we would have X is less than one-fifth
and X is greater than negative one-fifth.
We know this is and
because it's less than,
and remember for the second inequality,
we reverse the inequality
symbol and change the sign.
So now we know the open
interval of convergence
would be from negative one-fifth
to positive one-fifth, but we're not done.
We need to test the end points
to see if this series
would converge or diverge,
again, at the end points.
So this first test, X
equals negative one-fifth.
Notice when X is negative one-fifth,
we'd have the summation from
N equals one to infinity
of five times negative one-fifth,
raised to the Nth, divided
by N to the eighth,
simplifying, we'd have the
summation from N equals one
to infinity of, this would
be negative one to the Nth,
divided by N to the eighth.
Notice here we have an alternating series,
so let's apply the
alternating series test.
Notice ace of N, the non-alternating part,
would be one divided by N to the eighth
which is always greater than zero,
and now I'll find the
limit as N approaches
infinity of ace of N as well as C of ace
of N plus one is less
or equal to ace of N.
So the limit as N approaches N of infinity
of ace of N, or one
divided by N to the eighth,
is equal to zero and ace of N plus one
is less or equal to ace of N.
Notice as it increases, the fractions
would get smaller and
smaller, and therefore,
the series converges
at X equals negative one-fifth
by the alternating series test.
Now we'll see what happens
when X equals one-fifth.
Notice how the only difference here
is that we would have a
positive one-fifth here
so we'd have positive one to the Nth,
divided by N to the eighth, but of course,
one to the Nth is just one.
So we have one divided by N to the eighth,
which we should recognize would converge
by the p-series test with P equals eight.
So the series also converges
at the end point X
equals positive one-fifth
by the p-series test,
which means, the interval of convergence
would be not the open interval
from negative one-fifth
to positive one-fifth, but it
would be the closed interval
from negative one-fifth
to positive one-fifth.
So looking at our question here,
it says the series is
convergent at X equals
negative one-fifth, and the
left end point is included,
so the answer is yes, to X
equals positive one-fifth,
and the right end point is also included,
so we say yes again.
I hope you found this helpful.
