Last time, we talked about power
series, radii of convergence,
and the intervals of convergence.
This time we're gonna talk about how
you actually go about computing
radii of convergence and
interval of convergence.
And our big guns are the root and the
ratio tests—especially the ratio tests.
Root test comes in occasionally;
I find that the ratio test
is actually the more useful.
The point is, if you apply the
ratio test to a power series,
you're taking the limit of
a_(n+1)x^(n+1)/(a_nx^n).
And all of the factors of x cancel
except for 1, and that just pulls out.
So you really are just looking at the size
of x times the limit of a_(n+1)/a_n.
So if you can figure out
this limit, you're golden.
Likewise, if you apply the root test,
if you take the nth root of a_nx^n,
well the nth root of x^n is just x.
And the nth root of a_n, well that's
what you would study with the root test.
So the rule is that if you get a
number by either of the two tests,
and if you get the number by one test,
you're gonna get the same
number by the other test.
So if you get a number, let's call it L,
which is the limit of the
ratio of coefficients,
or L is the limit of the nth
root of the nth coefficient,
if that happens, then that gives
you the radius of convergence.
If L is 0, then the radius
of convergence is infinity,
because if this limit goes to 0,
then no matter how big x is,
this whole thing goes to 0.
And therefore the series converges.
If L is infinite, then the radius of
convergence is 0.
'Cause no matter how small x is,
this whole thing blows up,
and so the series diverges.
If L is a finite number, then the radius
of convergence is its reciprocal.
'Cause as long as the magnitude
of x is less than 1/L,
then this is gonna be less than 1.
If the magnitude of x is bigger than 1/L,
then this is gonna be bigger than 1.
And if the magnitude of x is equal to 1/L,
well, then you have to worry about
those special cases by hand.
Finally, sometimes you have a
situation where L doesn't exist.
You apply the root test or the ratio test
and you just don't get a consistent answer
in which case, oh well, it didn't work,
try something else.
So let's look at some examples.
Our first example is the sum of (n^3)x^n.
In this example, a_n is n^3.
And we compute by the ratio test the
limit of a_(n+1)/a_n,
it's the limit of ((n+1)/n)^3.
(n+1)/n goes to 1, so its cube goes to 1.
Since L is 1, R is the
reciprocal of 1, which is 1.
So the series diverges when
x is bigger than 1,
it converges when x is less than 1.
We have to check the cases
where x is 1 or -1 by hand.
When x is 1, we have just the sum of n^3.
Well that diverges,
'cause the terms get big.
When x is -1, you have ±n^3.
That also diverges.
So in this case, the series diverges
when it's on the borderline,
so the interval of convergence
is an interval of radius 1,
but that doesn't include the endpoints.
Our next example is the sum of n!x^n.
So a_n is n!,
L is the limit of (n+1)!/n!,
and that's n+1, and that goes to infinity,
so the radius of convergence is 0.
The only value of x for which
this converges is x = 0.
In which case all the terms but
the 0th term are 0,
so of course it converges.
The next example is the sum of (x^n)/n!.
Now, L is the limit of (1/(n+1)!)/(1/n!),
in other words, n!/(n+1)!.
That's 1/(n+1), that goes to 0.
So that means the radius
of convergence is infinite,
and the interval of convergence
is everything.
No matter how big x is,
this series converges.
And finally, let's look at
the sum of (x^n)/n.
Now, we take L as the limit of
(1/(n+1))/(1/n).
That limit is 1,
so radius of convergence is 1.
We look at what happens in
the boundary when x is -1,
this converges;
it's an alternating series.
When x is +1, it diverges.
So that means that the interval
of convergence is -1 to 1,
including -1, but not including 1.
