Math 414
Professor Lieberman
February 18, 2003
HOMEWORK #5 SOLUTIONS
Section 2.2
8. (a) We will show that A < B + ε for every ε > 0. So let ε > 0 be given. Then there are
numbers n1 and n2 such that |an − A| < ε/2 if n ≥ n1 and |bn − B| < ε/2 if n ≥ n2 . We
now choose n = max{n1 , n2 }. Then
ε ε
A − B = (A − an ) + (bn − B) + (an − bn ) < + + 0 = ε,
2 2
so A < B + ε as claimed.
(b) There are lots of examples. I like an = 0 and b + n = 1/n but there are many others.
11. (d) The limit exists and it’s zero. To see this, let m ∈ N be the number from Theorem
1.6.8 such that m > |r|, and let ε > 0 be given. Then Theorem 1.6.8 gives a number n∗
such that
1 mm
∗
n > max
, m+1 .
ε m!
If n ≥ n∗ , then
n
n
r
− 0 = |r| ≤ m m . . . m m . . . m .
n!
n!
n n−1
mm−1
1
But m/n ≤ m/n∗ so m/s ≤ 1 if s = m, . . . , 1, so
n
m
r
− 0 ≤ m m < ε.
n!
n∗ m!
√
√
(i) Let bn = n n and cn = n 2n. Then bn ≤ an ≤ cn and we know that (bn ) converges to 1
by Exercise 15 of Section 2.1 while (cn ) converges to 1 by combining Exercises 14 and 15
of Section 2.1 with Theorem 2.2.1(b). It then follows from the Squeeze Theorem that (an )
converges to 1.
13. It is not true
√ that the sequence√(bn ) must converge. For example, we could have an = 1/n
and bn = n. Then an bn = 1/ n, so an bn → 0, but (bn ) diverges.
Section 2.3
4. (a) Take bn = an .
(b) Take bn = (−1)n an .
8. First, we rewrite an = np−q bn with
bn =
sp + sp−1 n−1 + · · · + a0 n−p
.
tq + tq−1 n−1 + · · · + t0 n−q
From the limit theorems along with the equation sp tq 6= 0 (which is the same as sp 6= 0
and tq 6= 0), we have that (bn ) converges to sp /tq , which is not zero. It then follows from
1
2
Theorem 2.1.12, that bn is bounded, so Theorem 2.3.3(b) tells us that (an ) diverges to
infinity.
Section 2.4
1. One possibility is
(
n
an =
n/2
if n is even
if n is odd.
6. (a) We can write
1/2
1
+
,
2 2n + 1
so (an ) is decreasing, and it is bounded below by 1/2. Therefore it converges by Exercise
5 of Section 2.4 (which is the “decreasing” version of Theorem 2.4.2).
(e) Now we have to be a little trickier. Let’s look at an+1 /an :
an =
an+1
3n+1 1 + 32n
3 + 32n+1
=
=
an
1 + 32n+2 3n
1 + 32n+2
2n+1
2n+1
3
+3
2 · 32n+1
32n+2
≤
=
≤
≤ 1,
1 + 32n+2
1 + 32n+2
1 + 32n+2
so the sequence is decreasing. Since the terms are positive, it’s also bounded below, so it
converges by Exercise 5.
(f) Now we have
an+1
(n + 1)! 1 · 3 · · · · · (2n − 1)
n+1
=
=
≤ 1,
an
n!
1 · 3 · · · · · (2n + 1)
2n + 1
so the sequence is decreasing. Again the terms are positive, so it’s bounded below and
hence convergent.
7. (a) Now
1
> 0,
(n + 1)2
so the sequence is increasing. Exercise 2(t) from Section 1.3 says that an ≤ 2 − 1/n, so the
sequence is also bounded. Therefore it converges by Theorem 2.4.2. In addition an ≥ 1, so
Theorem 2.2.1(f) says that 1 ≤ A ≤ limn→∞ (2 − 1/n). This last limit is 2. In addition,
an ≥ 5/4 if n ≥ 2, so Theorem 2.2.1(f) also says that A ≥ 5/4.
an+1 − an =
Section 2.5
3. (a) an = (−1)n n/(n + 1) has exactly two accumulation points −1 and 1. (The problem
didn’t ask you to prove that these are all the accumulation points, but here’s a proof that
they are accumulation points anyway. First, given ε > 0, Theorem 1.6.8 says that there is
an integer n∗ such that n∗ > 1/ε. Then
2n∗
1
1
∗
|a2n − 1| = ∗
− 1 = ∗
≤ ∗ < ε,
2n + 1
2n + 1
n
a2n∗ ∈ S, and a2n∗ 6= 1. Similarly |a2n∗ +1 − (−1)| < ε, a2n∗ +1 ∈ S, and a2n∗ +1 6= −1. Hence
−1 and 1 are accumulation points. Showing that there aren’t any more accumulation points
3
is similar but a lot of work!)
(b) Same as (a).
(c) S = (0, 1) ∪ {2}.
(d) S = {−1} ∪ (0, 1) ∪ {2}.
4. (a) If s0 is an accumulation point for S, then, for each positive integer n, there is a point
an ∈ S with an 6= s0 and |an − s0 | < 1/n (this is just the point corresponding to ε = 1/n).
To see that (an ) converges to s0 , let ε > 0 be given. Then Theorem 1.6.8 gives a natural
number n∗ such that n∗ > 1/ε. If n ≥ n∗ , then
1
1
|an − s0 | < ≤ ∗ < ε,
n
n
so an → s0 .
Conversely, if there is a sequence (an ) in S with an 6= s0 for all n and an → s0 , then,
given ε > 0, there is a natural number n∗ such that |an − s0 | < ε if n ≥ n∗ . It follows that
an∗ ∈ S, an∗ 6= s0 and |an∗ − s0 | < ε. Hence s0 is an accumulation point of S.
(b) It’s necessary because the set S = {2} has no accumulation points but the sequence
an = 2 for all n is a sequence of points in S such that an → s0 .