MATH 3A03 Assignment #2
Due: Friday, October 2, in class
1. Let A and B be nonempty bounded subsets of R. Define A + B to be
the set
{a + b : a ∈ A and b ∈ B}.
Show that A + B is upper bounded and that sup (A + B) = sup A +
sup B. (Note that A + B is also lower bounded and inf(A + B) =
inf A + inf B.)
Solution: Let M = sup A and N = sup B. First show that M + N
is an upper bound of the set A + B. Let r ∈ A + B. Then there is
some a ∈ A and b ∈ B with r = a + b. Since a ≤ M and b ≤ N then
r = a + b ≤ M + N . This establishes that M + N is an upper bound
of the set.
To show that M + N is the least upper bound of A + B, let s < M + N .
We will show that s is not an upper bound of A + B to conclude the
solution. What we need to do is to show that there is a ∈ A and b ∈ B
such that s < a + b (since this would show that s can’t be greater than
or equal to all of the members of A + B). Of course, we will need to
use that M and N are the suprema of A and B respectively, somehow.
Since s < M + N then s − N < M and so s − N is not an upper bound
for A. So, there is some a ∈ A with s − N < a (if not, then s − N
would also be an upper bound for A that is smaller than M ). Then
s < a + N and so s − a < N . A similar argument now gives that there
is some b ∈ B with s − a < b and from this we conclude that s < a + b,
as required.
2. Use the formal definition of a limit to prove that the following sequences
converge:
(a) sn = 5 − 4−n
Solution: We will show that the limit of {5 − 4−n } is equal to 5
by using the formal definition of the limit. Let > 0 be given.
This part of the solution can be considered as “rough work” and
will be used to find a suitable integer N that satisfies the condition
1
in the definition of the limit for this given value of . We want to
ensure the |sn − 5| < by taking n to be big enough. Note that
|sn − 5| = |5 − 4−n − 5| = |4−n | = 4−n
and so it suffices to ensure that 4−n is less than . To make 4−n < we need to have that 4n > 1/ or n > − log4 (). So, setting N to
be any natural number greater than − log4 () will work.
Now, for the formal part of the solution. For > 0, let N be a
natural number that is greater than − log4 (). Then if n ≥ N , we
have that
|sn − 5| = |5 − 4−n − 5| = |4−n | = 4−n < 4log4 () = ,
as required.
n+4
(b) sn = 2
3n + 11
Solution: This limit is similar to the one that was worked out
during the lectures. The strategy is to come up with a simpler
expression in n that is greater than or equal to sn and that we can
bound below a given . Note that there are many ways to come
up with a suitable expression. By considering the terms of this
sequence, we can see that the limit of it will be equal to 0. So,
given and > 0 we want to find out how big n must be to ensure
that |sn − 0| < .
(rough work) We can use that sn > 0 for all n ∈ N to remove
the absolute value sign in |sn − 0|. Next, using that 3n2 + 11 >
n+4
n+4
<
. As long as n ≥ 4, we have
3n2 , we get that
2
3n + 11
3n2
that n + 4 ≤ 2n. Combining this observation with the previous
inequality, we get that
|sn − 0| <
n+4
2n
2
≤ 2 =
.
2
3n
3n
3n
So, as long as 2/(3n) < , we will also have that |sn − 0| < .
Solving the inequality 2/(3n) < for n, we get that n > 2/(3).
The conclusion is that setting N to be a natural number greater
than 2/(3) and 4 will work.
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(back to the solution proper) Given > 0, let N be any natural
number greater than max{4, 2/(3)}. Then if n ≥ N ,
|sn − 0| =
n+4
2
<
< ,
2
3n + 11
3n
as required. Note that the last inequality holds, since n > 2/(3)
implies that 2/(3n) < .
3. Use the formal definition of a limit to prove that the following sequences
diverge:
√
(a) sn = 3 n
Solution: Note that this sequence is unbounded and so by a theorem proved in class (Theorem 2.11), we know
√ that this sequence
is divergent. To prove directly that limn→∞ 3 n = ∞, we√need to
show that for all M ∈ R, there is some N ∈ N such that 3 n ≥ M
whenever n ≥ N . If we set N to be any natural number that is
greater than M 2 /9, then if n ≥ N , we have that
p
√
3 n > 3 M 2 /9 = M,
as required.
nπ
(b) sn = cos
3
Solution: Observe that this sequence is periodic, since for all n,
sn = sn+6 . Furthermore, the first six terms of this sequence are
1/2, −1/2, −1, −1/2, 1/2, 1
and this pattern of terms repeats indefinitely. We see that this
sequence does not approach any fixed real number L and so it
diverges.
To prove this claim, let L ∈ R and show that the sequence doesn’t
converge to L. First consider the case that L 6= 1. Let = |L−1|/2
(or any positive number that is less than |L − 1|. Note s6k = 1 for
all k ∈ N and so for arbitrarily large n we have that |sn − 1| > .
Thus, for every N ∈ N, there is some n ≥ N with |sn − 1| > .
According to the definition of the limit, this establishes that {sn }
does not converge to L.
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The remaining case is when L = 1 and it almost identical to
the previous case; just use the number -1 in place of 1 in the
argument from the previous paragraph to show that the sequence
can’t converge to any number L that is not equal to -1, including
the case L = 1.
4. Supppose that lim sn = 0 and that the sequence {tn } is bounded (but
n→∞
not necessarily convergent). Prove that lim (sn tn ) = 0. Provide a
n→∞
counterexample to show that the assumption that the sequence {tn } is
bounded is necessary.
Solution: It is worthwhile to compare the proof that the product of two
convergent sequences is convergent with the following solution. In this
problem we don’t know that {tn } is convergent, only that it is bounded.
The other element that will be used is that the sequence {sn } converges
to 0.
Let M > 0 be a bound for {tn }. This means that |tn | ≤ M for all n.
To show that lim (sn tn ) = 0, let > 0 and find N so that whenever
n→∞
n ≥ N , |sn tn − 0| < . Since {sn } converges to 0 then there is some
natural number N so that whenever n ≥ N , |sn − 0| < /M . But then
for any n ≥ N we have that
|sn tn − 0| = |sn ||tn | ≤ |sn |M < (/M )M = ,
as required.
If the sequence {tn } is not bounded then the conclusion won’t necessarily follow. For example, let {sn } = {1/n} and let {tn } = {n2 }.
Then {sn } converges to 0, but the sequence {sn tn } = {n} is a divergent
sequence.
1
= 0. Does the converse hold,
n→∞ sn
5. Prove that if lim sn = ∞ then lim
n→∞
1
i.e., if lim
= 0 then does lim sn = ∞?
n→∞ sn
n→∞
Solution: To show that limn→∞ s1n = 0, let > 0 be given. Let M = 1/.
(if is small then M will be big.) Since we know that lim sn = ∞
n→∞
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then for this value of M , there is some integer N such that sn > M
whenever n ≥ N . But then, whenever n ≥ N we have that
|1/sn − 0| = 1/sn < 1/M = .
This establishes that {1/sn } converges to 0.
For the other direction, consider the sequence {sn } = {(−1)n n}. This
is a divergent sequence, since it is not bounded. But, using the Squeeze
Theorem we see that the sequence {1/sn } = {(−1)n /n} converges to 0
(since the terms of this sequence lie between the terms of the sequences
{−1/n} and {1/n} and these two sequences converge to 0). Note that if
in addition, we know that all of the terms of the sequence are positive,
then it does follow that lim sn = ∞.
n→∞
6. Suppose that lim sn = ∞ and {tn } is a sequence such that for all
n→∞
n ∈ N, sn ≤ tn . Prove that lim tn = ∞.
n→∞
Solution: To show that lim tn = ∞, let M ∈ R and show that there
n→∞
is some N ∈ N such that tn ≥ M whenever n ≥ N . Using that
lim sn = ∞, we have that for this given value of M there is some
n→∞
N ∈ N such that whenever n ≥ N , sn ≥ M . But then whenever
n ≥ N , we have that
tn ≥ sn ≥ M,
as required.
Note that one can also use the result from the previous problem as
follows. Since lim sn = ∞ then we may assume that the terms of {sn }
n→∞
and {tn } are all positive (since they will all eventually be greater than
1
= 0. Using
0). From the previous question, we conclude that lim
n→∞ sn
the inequality sn ≤ tn we get that 0 ≤ 1/tn ≤ 1/sn and so by the
1
Squeeze Theorem, that lim
= 0.
n→∞ tn
By the remark at the end of the previous solution, we can then conclude that lim tn = ∞ since we have assumed that the tn are all
n→∞
non-negative.
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Supplementary problems from the textbook
(not to be handed in)
2.4.2, 2.4.3, 2.4.5, 2.4.11, 2.5.3, 2.5.5, 2.5.6, 2.6.1, 2.6.2, 2.7.4, 2.8.1, 2.8.5,
2.8.7, 2.8.9.
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