Problem 0.1. Determine the interval of convergence for the following series:
∞
X
xn f (n)
n!
n=0
where
f (n) =
(
1 if x is even
2
1/2n if x is odd.
We can try the ratio test, but there’s lots of weird ups and downs in the
terms. We can either use the comparison test (using the fact that 2n12 is smaller
than 1) or use the root test. For practice, let’s use the root test:
r
xn f (n)
p n!
x n f (n)
≤ √
n
n!
x
≤ √
n
n!
n
You can rewrite this logarithmically as
x
e
1
n (log 1+...+log n)
which goes to zero as n → ∞ because log n goes to infinity, so the averages do
too.
Problem 0.2. Find a power series that has (2, 8] as its interval of convergence.
The trick here is to determine the center of the interval of convergence- that
is x = 5. So you want to have a power series with (x − 5) terms. You want the
radius of convergence to be 3 so it would be helpful if the denominator grew at
a rate of about 3n . What about the endpoints? The easiest way to do this is
to make the series absolutely divergent for |x − 5| = 3 but alternating at x = 8.
One series with all of these properties is
∞
X
(−1)n (x − 5)n
.
n1/2 3n
n=1
1