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Math 1B Discussion Section SOLUTIONS
Rob Bayer
October 16, 2007
You should work on the following problems in groups of 3 or 4. Try to get through as many as you can,
but you aren’t expected to finish everything. Instead, you should make sure everyone in your group knows
how to solve all the problems and not just the answers.
- Find the radius and interval of convergence for each of the following power series:
(a)
∞ ∑
n=
x
n
n^2
(b)
∞ ∑
n=
(x − 2)
n
n n
We’ll use the root test: limn→∞
n
(x−2)n nn
∣ = lim
|x− 2 | n = 0, which is less than 1 for all x, so
R = ∞ and I = (−∞, ∞).
(c)
∞ ∑
n=
x
n ln(1 +
n
) We’ll use the ratio test:
lim n→∞
x n+ ln(1 + 1 n+
x n ln(1 +
1 n
LH = lim
x
1 1+ 1 n+
1 (n+1)^2
1 1+ 1 n
1 n^2
= lim
x
1 (n+1)^2 +(n+1) 1 n^2 +n
= lim
x
n
2
(n + 1)^2 + n + 1
= |x|
So we just want |x| < 1. That is, − 1 < x < 1. Now we test the endpoints:
(x = −1):
n ln(1 +
1 n ) is an alternating series with bn = ln(1 +
1 n ) and thus converges since
bn → 0 and is decreasing.
(x = 1):
ln(1 +
1 n
ln(
n+ n
ln(n + 1) − ln(n), which is a telescoping series. We write
out a few terms to see that
sn = ln 2 − ln 1 + ln 3 − ln 2 + ln 4 − ln 3 + · · · + ln(n + 1) − ln n = − ln 1 + ln(n + 1) = ln(n + 1)
Then lim sn = ∞ and thus the series does not converge.
Now that we’ve tested the endpoints, we can see that R = 1 and I = [− 1 , 1)
(d)
∞ ∑
n=
(3x − 1)
n
n 4 n
As usual, we do the ratio test:
lim n→∞
(3x − 1) n+
(n + 1) n+
n 4 n
(3x − 1) n
= lim
(3x − 1)n
4(n + 1)
| 3 x − 1 |
3 |x −
1 3
So we want 3|x − 1 / 3 |/ 4 < 1, which is the same as |x − 1 / 3 | < 4 /3, so R = 4/3, and the center of
our interval is a = 1/3. To find the endpoints, we re-write this inequality as − 4 / 3 < x− 1 / 3 < 4 /3,
and add 1/3 to everything to get − 1 < x < 5 / 3
(x = −1): The series is then
∑ (^) (−4)n
n 4 n^
∑ (^) (−1)n
n
, which converges by the alternating series test.
(x = 5/3): We have
4 n
n 4 n^
1 n , which is divergent by the p-series test.
Therefore, I = [− 1 , 5 /3) and R = 4/ 3
- Given the interval of convergence for
anx n , can you determine the radius? How about the other
way around?
Given an interval, the radius is just half the length of it, which is really easy to calculate (Note: this
is why we call it the radius of convergence–just like the radius of a circle is half the distance across,
the radius of an interval is half its length). Given just a radius, you don’t know anything about the
endpoints and thus can’t determine the interval.
- True/False: For those that are true, explain why. For those that are false, give a counterexample.
(a) If
an(−3)
n converges, then
an 2
n converges.
True. Consider the power series
anx
n
. Then we know there is some number R such that
it converges whenever |x| < R and diverges whenever |x| > R. Note that
an(−3)
n being
convergent is the same as saying that this power series converges at x = −3. Then we know that
R must be at least 3. Then we must have that
anx
n converges whenever |x| < 3 ≤ R, so in
particular it converges at x = 2, which is just
an 2
n
(b) If
an(−3) n converges, then
an 3 n converges.
False.
x n
n 3 n^
converges at x = −3, but diverges at x = 3.
(c) If
anx n has a positive radius of convergence, then lim an = 0
False. Consider
x n , which has radius of convergence 1. But each an is just 1, so the limit is
too.
- For each n, let fn(x) =
sin(nx) n^2
. Show that
fn(x) converges for all x. Is the same true of the series
of derivatives,
f ′ n (x)?
a)
| sin(nx)| n^2
1 n^2 , so by the comparison test,
| sin(nx)| n^2 converges and thus
sin(nx) n^2 converges abso-
lutely (and thus converges).
b) f ′ n
n cos nx n^2
cos(nx) n
. Note for x = 0, this is just
1 n
, so the series
f ′ n diverges at x = 0 (actually,
any multiple of 2π would have worked here).
By now, you’ve seen that you can differentiate power series term-by-term and the result still converges
on the same interval (except maybe the endpoints). This example shows that that is a special property
of power series, and is not true of a series made out of just any functions.
