Welcome to part one of Power Series.
The goals of this video are
to define a power series,
and also to determine the radius
and interval of convergence
when centered about x equals zero.
Part two will show how to solve problems
when it's not centered at zero.
A power series in one
variable is an infinite series
in the form as we see here,
written using sigma notation
as well as expanded.
This power series here is centered at c.
And a sub n represents the
coefficient of the nth term
and the value of x will vary around c.
And if the power series is
centered at c equals zero,
the power series would
simplify to the series
that we see here.
The power series will
converge for some values of x
and may diverge for others.
So our main goal for this video
is to determine the interval
for which the series converges.
All power series will
converge at x equals c
where it's centered.
If c is not the only convergent point,
then there is always a number
R, often called the radius
that is greater than zero or
less than equal to infinity
such that the series will converge
whenever the distance between
x and c is less than R,
and diverges when the
distance between x and c
is greater than R.
And this number R is called
the radius of convergence
of the power series.
So this is how we'll approach finding
the interval of convergence.
If R is equal to zero, the power
series only converges at c.
If R equals infinity then
the power series converges
for all values of x.
And if we do find a specific value for R,
R tells us the series will
converge on the open interval
from c minus R to c plus R.
Notice this is an open interval,
however to determine the
interval of convergence,
we do have to check the
two endpoints to determine
if it converges or diverges
at those two points.
And this information will give us
the interval of convergence.
So let's go ahead and give it a try.
Here we have a power series.
And we'll start by finding
the radius of convergence.
So we need to apply one
of the convergence tests
to determine for which values of x
this series will converge.
And since we have an
exponential in our formula,
we'll go ahead and apply the
ratio test as we see here.
So we'll take the limit
as n approaches infinity
of the absolute value of a sub n plus one
divided by a sub n.
So the numerator will be a sub n plus one.
So we'll replace n with n plus one.
And our denominator will
be a sub n which is just
n times x to the nth.
Now we're not including
negative one to the power of n
because we are taking the absolute value.
So now we'll go ahead and simplify this.
So the n plus one over
n does not simplify out.
But here we have x to the n
plus one divided by x to the n.
There's one more factor
of x in the numerator,
so we're left with an x in the numerator.
Now as n approaches infinity,
n plus one divided by n approaches one,
so this limit is equal to
the absolute value of x.
Remember using the ratio test,
this limit must be less than one.
So in order for this
power series to converge,
the absolute value of x
must be less than one.
This tells that the radius of
convergence is equal to one.
And since this power series
is centered at x equals zero,
that means the open
interval of convergence
will be zero minus one all
the way to zero plus one,
which means we know that it converges
from negative one to positive one.
But to determine the true
interval of convergence,
we do have to test
whether this power series
converges or diverges at the endpoints
or negative one and positive one.
So we'll do that by replacing x
with negative one and positive one
and see if that series
converges or diverges.
Let's go ahead and do
that on the next screen.
So here's the given power series.
We know for sure this
power series converges
on the open interval from
negative one to positive one,
but now we need to see what's happening
at negative one and positive one.
So when x equals negative
one, we'll have the summation
from one to infinity of
negative one to the n
times n times negative one to the n again.
Well, negative one to the n
times negative one to the n
will always be positive one.
So this leaves us with the summation
from n equal one to infinity of n.
Well, using the nth term divergent test,
if we take the limit as
n approaches infinity
of a sub n, which in this case, is just n,
this does not equal zero,
therefore the series diverges
at x equals negative one.
So we cannot include negative one
as part of the interval of convergence.
Now let's see what
happens when x equals one.
If x equals one, we'll
have one to the nth.
And even though this
series does alternate,
if we take the limit of a sub n,
we still have the limit as
n approaches infinity of n
which, again, doesn't equal zero.
So by the nth term divergent test,
this series also diverges at x equals one.
So since this series diverges
at the endpoints of negative
one and positive one,
this open interval is the
interval of convergence.
Let's go ahead and try another one.
Notice this power series
also has an exponential part,
letting us know that the ratio
test is probably going to be
a good choice to determine
the radius of convergence.
So let's go ahead and
apply the ratio test again.
We'll have the limit as
n approaches infinity
of a sub n plus one,
which will be x to the n plus
one divided by n plus one.
And then instead of dividing by a sub n,
we'll multiply by the reciprocal.
So instead of x to the n divided by n,
we'll have n divided
by x to the power of n.
Now, let's go ahead and simplify this.
Notice we have one more
factor of x in the numerator,
so we'll have x times n
divided by n plus one.
So as n approaches infinity,
we have n over n plus one.
This approaches one,
so this limit is equal
to the absolute value of x,
which must, again, be less than one.
So this tells us the radius of
convergence is equal to one.
And again, since this power series
is centered at x equals zero,
the open interval of convergence will be
from zero minus one to zero plus one,
or from negative one to one.
So now we'll test this power series to see
if it converges or diverges
at negative one and one,
just like we did on the last problem.
Let's first let x equal negative one.
So we'll have negative one to the n
times negative one to the n all over n.
And again, negative one to the
n times negative one to the n
equals positive one.
So we have one over n.
And we should recognize
this as a harmonic series
which we know diverges
from the p series test
where p is equal to positive one.
Now let's see what
happens when x equals one.
So we'll have negative one to the n
times one to the nth, divided by n.
Well one to the nth will always be one.
And we should recognize this
as the alternating harmonic
series which does converge.
Let's go ahead and apply
the alternating series test.
So we first need to check to make sure
the limit of a sub n is equal to zero
as n approaches infinity.
So we'll have the limit
as n approaches infinity
of one over n, which does equal zero.
That's the first check.
And then we need to make
sure that zero is less than
a sub n plus one which
is less than a sub n.
So a sub n plus one would
be one over n plus one,
and a sub n, of course,
is just one over n.
Well, this fraction here
has larger denominators,
so therefore, it will always
be less than one over n,
so the second condition is also met.
So by the alternating series test,
the series converges at x equals one.
So the series diverged at negative one,
so we cannot include that endpoint,
but it does converge at x equals one,
so we do have to include one
in the interval of convergence.
So the interval of convergence will be
from negative one to positive one,
where it's open on negative
one and closed on positive one.
That's going to do it
for this first video.
We'll take a look at some
additional examples in part two.
Thank you for watching.
