Math 317 HW #4 Solutions
1. Exercise 2.2.5. Let [[x]] be the greatest integer less than or equal to x. For example, [[π]] = 3
and [[3]] = 3. Find lim an and supply proofs for each conclusion if
(a) an = [[1/n]].
Answer. In this case, lim an = 0. To see this, let > 0. Then, for all n ≥ 2, we know
that n1 ≤ 12 < 1. On the other hand, since n > 0, certainly n1 > 0. Therefore
0 < an < 1,
so [[an ]] = 0. Hence, for all n ≥ 2, we have that
|an − 0| = |0 − 0| = 0 < .
Since the choice of > 0 was arbitrary, this implies that lim an = 0.
(b) an = 10+n
.
2n
Answer. Again, lim an = 0. To see this, let > 0. Then, for all n > 10, we have that
5 1
5
1
10 + n
= + <
+ = 1.
2n
n 2
10 2
On the other hand, certainly 0 < 10+n
2n for all n. Therefore, when n > 10 we have that
an = 0, and so
|an − 0| = |an | = 0 < .
Since the choice of > 0 was arbitrary, we conclude that lim an = 0.
√
2. Exercise 2.2.7. Informally speaking, the sequence n “converges to infinity.”
(a) Imitate the logical structure of Definition 2.2.3 to create a rigorous definition for the
√
mathematical statement lim xn = ∞. Use this definition to prove lim n = ∞.
Answer. Here’s a definition for lim xn = ∞:
A sequence (xn ) converges to ∞ if, for every positive number M , there exists N ∈ N
such that whenever n ≥ N it follows that |xn | > M .
√
Now we prove that lim xn = ∞:
Proof. Let M > 0. By the Archimedean property, there exists N ∈ N such that N > M 2 .
Then, whenever n ≥ N , we have that
√
√
√ √
n = n ≥ N > M 2 = M.
√
Therefore, since the choice of M > 0 was arbitrary, it follows that lim n = ∞.
(b) What does your definition in (a) say about the particular sequence (1, 0, 2, 0, 3, 0, 4, 0, 5, 0, . . .)?
Answer. According to the above definition, the given sequence does not converge to ∞
since, for any n ∈ N, x2n = 0 6> M no matter what the choice of M > 0.
3. Exercise 2.3.4. Show that limits, if they exist, must be unique. In other words, assume
lim an = `1 and lim an = `2 , and prove that `1 = `2 .
1
Proof. Assume lim an = `1 and lim an = `2 . By swapping the names `1 and `2 if necessary,
assume `2 6= 0 (if this can’t be done, then we already have `1 = 0 = `2 ). Then, by the
Algebraic Limit Theorem,
an
lim an
`1
1 = lim 1 = lim
=
= ,
an
lim an
`2
which implies `1 = `2 .
An Alternative Proof. Let > 0. Then, since lim an = `1 , there exists N1 ∈ N such that, for
all n ≥ N1 ,
|an − `1 | < /2.
Likewise, since lim an = `2 , there exists N2 ∈ N such that, for all n ≥ N2 ,
|an − `2 | < /2.
Therefore, if N = max{N1 , N2 } and n ≥ N , we have that
|`1 − `2 | = |`1 − an + an − `2 | ≤ |`1 − an | + |an − `2 | < /2 + /2 = .
Therefore, since our choice of > 0 was arbitrary, we see that |`1 − `2 | is smaller than all
positive numbers; since it is non-negative, this implies |`1 − `2 | = 0, meaning that `1 = `2 .
4. Exercise 2.3.7.
(a) Let (an ) be a bounded (not necessarily convergent) sequence, and assume lim bn = 0.
Show that lim(an bn ) = 0. Why are we not allowed to use the Algebraic Limit Theorem
to prove this?
Proof. Since (an ) is bounded, there exists M > 0 such that |an | < M for all n ∈
{1, 2, 3, . . .}. Moreover, since lim bn = 0, there exists N ∈ N such that, for all n ≥ N ,
|bn − 0| < /M
or, equivalently, |bn | < /M .
Therefore, whenever n ≥ N , we have that
|an bn − 0| = |an bn | = |an ||bn | ≤ M |bn | < M
= .
M
Since the choice of > 0 was arbitrary, this implies that lim(an bn ) = 0.
The Algebraic Limit Theorem would imply that
lim(an bn ) = (lim an ) (lim bn ) = (lim an ) (0) = 0
whenever (an ) converges but, since we are not guaranteed this, the Algebraic Limit
Theorem does not necessarily apply.
2
(b) Can we conclude anything about the convergence of (an bn ) if we assume that (bn ) converges to some nonzero limit b?
Answer. No, we cannot conclude anything. If (an ) converges to some a, then of course
the Algebraic Limit Theorem implies that (an bn ) → ab. However, when (an ) does not
converge, the sequence (an bn ) may not converge.
For example, when an = (−1)n and bn = 1, then (an ) is bounded and (bn ) converges (to
1), but
(an bn ) = (−1, 1, −1, 1, −1, 1, . . .),
which does not converge.
(c) Use (a) to prove Theorem 2.3.3, part (iii), for the case when a = 0.
Proof. Remember that Theorem 2.3.3 (iii) says that lim(an bn ) = ab when lim an = a and
lim bn = b. We can prove this easily enough, now: since (bn ) converges, (bn ) is bounded.
Therefore, by (i) (switching the roles of an and bn ), the sequence (an bn ) converges to 0,
which of course is equal to ab since a = 0.
5. Exercise 2.3.11. Show that if (xn ) is a convergent sequence, then the sequence given by the
averages
x1 + x2 + . . . + xn
yn =
n
also converges to the same limit.
Give an example to show that it is possible for the sequence (yn ) of averages to converge even
if (xn ) does not.
Proof. Suppose lim xn = L. Then note that
n
yn − L =
n
1X
1X
n
xi − L =
xi − L
n
n
n
i=1
i=1
" n
#
1 X
=
xi − nL
n
i=1
" n
#
n
X
1 X
=
xi −
L
n
=
1
n
i=1
n
X
(1)
i=1
(xi − L).
i=1
Now, suppose > 0. Since (xn ) → L, there exists N1 ∈ N such that, for all n ≥ N1 ,
|xn − L| < /2.
In particular, this will imply that for all n ≥ N1
n
n
1 X
1 X n − N1 |xi − L| <
=
≤ ,
n
n
2
n 2
2
i=N1 +1
i=N1 +1
3
(2)
since
n−N1
n
≤ 1.
By the Archimedean Property, there exists N2 ∈ N such that
N
N2 >
1
2X
|xi − L|.
i=1
This implies that for any n ≥ N2 ,
N1
N1
1X
1 X
|xi − L| <
|xi − L| ≤
n
N2
2
(3)
i=1
i=1
Therefore, if N = max{N1 , N2 } and n ≥ N , we have that
n
n
1X
1 X
(xi − L) ≤
|xi − L|
|yn − L| = n
n
i=1
(4)
i=1
by (1) and the triangle inequality. In turn,
N1
n
n
1X
1X
1 X
|xi − L| =
|xi − L| +
|xi − L| < + = n
n
n
2 2
i=1
i=1
(5)
i=N1 +1
by (2) and (3).
Therefore, combining (4) and (5) yields
|yn − L| < .
Since the choice of > 0 was arbitrary, we conclude that (yn ) → L.
Example. Let xn = (−1)n . Then the sequence (xn ) definitely does not converge. Let’s write
the first few terms in the sequence (yn ):
y1 =
y2 =
y3 =
y4 =
y5 =
x1
−1
=
= −1
1
1
x1 + x2
−1 + 1
=
=0
2
2
x1 + x2 + x3
−1 + 1 − 1
1
=
=−
3
3
3
x1 + x2 + x3 + x4
−1 + 1 − 1 + 1
=
=0
4
4
x1 + x2 + x3 + x4 + x5
−1 + 1 − 1 + 1 − 1
1
=
=−
5
5
5
..
.
In general,
(
0
yn =
1
− 2n+1
if n is even
if n is odd.
It is straightforward to check that this sequence converges to 0.
4