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The University of Sydney
School of Mathematics and Statistics
Solutions to Tutorial 3 (Week 4)
MATH2962: Real and Complex Analysis (Advanced)
Semester 1, 2015
Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/
Lecturer: Florica C. Cˆırstea
Questions marked with * are more difficult questions.
Material covered
(1) Definition and properties of limits, limit inferior and limit superior;
(2) Limits and the limit laws;
(3) Inequalities such as the arithmetic-geometric mean inequality.
Outcomes
This tutorial helps you to
(1) be able to work with inequalities, limits and limit inferior/superior;
(2) have solid foundations in the more formal aspects of analysis, including a knowledge of
precise definitions, how to apply them and the ability to write simple proofs;
Summary of essential material
Monotone sequences in R always have a proper or improper limit. A proper limit means that
the limit exists in R, and an improper limit means the sequence diverges to +∞ or −∞.
Using that fact we define limit inferior and limit superior of an arbitrary sequence (xn ) in
R as follows. For n ∈ N set
an := inf{xn , xn+1 , xn+2 , xn+3 , . . . } = inf xk
k≥n
bn := sup{xn , xn+1 , xn+2 , xn+3 , . . . } = sup xk
k≥n
Because {xn+1 , xn+1 , xn+2 , xn+3 , . . . } ⊆ {xn , xn+1 , xn+2 , xn+3 , . . . }, properties of infimum and
supremum imply that (an ) is increasing and (bn ) is decreasing. Hence
lim inf xn := lim an = lim (inf xk ),
n→∞
n→∞
n→∞ k≥n
lim sup xn := lim bn = lim (sup xk )
n→∞
n→∞
n→∞ k≥n
exist either as a proper or as an improper limit. Facts about limit inferior and limit superior:
• lim inf xn ≤ lim sup xn ;
n→∞
n→∞
• lim inf xn = lim sup xn if and only if lim xn exists (as a proper or improper limits).
n→∞
n→∞
n→∞
If that is the case all three are equal.
c 2015 The University of Sydney
Copyright 1
We call a sequence (xnk )k∈N or simply (xnk ) a subsequence of (xn ) if (nk ) is strictly increasing
and nk → ∞ (the latter is automatic if (nk ) is strictly increasing, but we still make that explicit).
Subsequences have the following properties:
• The limit inferior of (xn ) coincides with the smallest accumulation point of (xn ); or −∞.
• The limit superior of (xn ) coincides with the largest accumulation point of (xn ), or +∞;
• Every bounded sequence has a convergent subsequence (Theorem of Bolzano-Weierstrass).
Questions to complete during the tutorial
3(−1)n n2
, n ≥ 0.
1. Consider the sequence xn = 2
n −n+1
(a) Sketch the graph of (xn ). It helps to look at monotonicity properties of |xn | by
multiplying xn by n2 /n2 and doing a completion of squares.
Solution: Multiplying xn by n2 /n2
|xn | =
3n2
3
=
1
n2 − n + 1
1− n +
=
1
n2
3
( n1
−
1 2
)
2
+
3
4
for all n ≥ 1. Since ( n1 − 12 )2 is increasing for n ≥ 2 it follows that |xn | is decreasing
for n ≥ 2. Moreover, x0 = 0, x1 = −3, x2 = 4 and x3 = −27/7. Hence the graph
looks as follows:
xn
3
1
2
3
4
5
6
7
8
9
10 11 12 13 14
N
−3
(b) Find an = inf k≥n xk and bn = supk≥n xk .
Solution: From the graph in the previous part and the values computed we have




−27/7
for
n
=
0,
1,
for n = 0, 1,

4
an = xn+1
bn = xn
for n ≥ 2 even,
for n ≥ 2 even,




xn
for n ≥ 3 odd,
xn+1 for n ≥ 3 odd.
(c) Hence compute lim inf n→∞ xn and lim supn→∞ xn .
Solution: Because
3
1−
1
n
+
1
n2
→3
We get lim inf n→∞ xn = limn→∞ x2n+1 = −3 and lim supn→∞ xn = limn→∞ x2n = 3.
2
2. Compute the limit
the definition, and
accumulation. The
(
n
(a) xn =
1/n
inferior and limit superior of the following sequences using directly
then using the fact that they are the smallest and largest point of
latter method is the one commonly used.
n even
n odd
Solution: We clearly have
1
= 0.
k≥n k
an = inf xk = inf
k≥n
for all n ≥ 0. Hence
lim inf xn = lim an = lim 0 = 0.
n→∞
n→∞
n→∞
Since the sequence is not bounded from above lim supn→∞ xn = ∞. We illustrate
the above using the graph of (xn ):
xn
1
2
3
4
5
6
7
8
9 10 11 12
N
We now identify the possible limits of convergent subsequences. From the definition
of the sequence we see that limn→∞ x2n = ∞ and limn→∞ x2n+1 = 0. Hence, any
subsequence of (xn ) which has a limit will either tend to ∞ or to 0 as n → ∞.
Hence, lim inf n→∞ xn = 0 and lim supn→∞ xn = ∞. That is the same as we
obtained before.
1
(b) 1 + (−1)n .
n
1
Solution: Note that is decreasing. Hence we have
n

1

1 −
if n is odd
n
an = inf xk =
1
k≥n

1 −
if n is even
n+1
and

1

1 +
if n is odd
n+1
bn = sup xk =
1

k≥n
1 +
if n is even
n
3
This is best seen from the graph of (xn ):
xn
3
1
2
3
4
5
6
7
8
9
10 11 12 13 14
N
We see that an → 1 and also bn → 1. Both coincide with the limit of (xn ). Rather
than compute the limit inferior and limit superior we could compute the limit, and
conclude the latter are the same.
3. Compute the limit inferior and limit superior of the following by using the fact that they
are the smallest and largest point of accumulation.

n

n even
(−1)n/2
n+1
2
(a) xn =
n −1

 2
n odd
2n + 1
Solution: We identify the accumulation points of (xn ) by looking for convergent
subsequences. We note that
x4n → 1
x4n+2 → −1
1
2
1
→
2
x4n+1 →
x4n+3
as n → ∞. Hence any convergent subsequence has either limit 1, −1 or 1/2. As
the smallest and largest are −1 and 1, respectively, we have
lim inf xn = −1,
n→∞
(b) sn =
n
X
lim sup xn = 1.
n→∞
(−1)k
k=0
Solution: We have
s0
s1
s2
s3
s4
..
.
=1
=1−1=0
=1−1+1=1
=1−1+1−1=0
=1−1+1−1+1=1
sn = 0
sn = 1
if n is odd
if n is even.
One should really do this by induction, observing that s0 = 1 and sn+1 = sn +
(−1)n+1 . Hence lim inf n→∞ sn = 0 and lim supn→∞ sn = 1.
4. Let (xn ) and (yn ) be bounded sequences in R.
4
(a) Let (xn ) and (yn ) be sequences in R. Prove that
lim sup(xn + yn ) ≤ lim sup xn + lim sup yn .
n→∞
n→∞
n→∞
Clearly xℓ + yℓ ≤ sup xk + sup yk for all ℓ ≥ n. This means that
Solution:
k≥n
k≥n
supk≥n xk + supk≥n yk is an upper bound for xℓ + yℓ for all ℓ ≥ n. By definition of
a supremum (every upper bound is larger or equal than the supremum)
sup(xℓ + yℓ ) ≤ sup xk + sup yk
ℓ≥n
k≥n
k≥n
for all n ∈ N. By definition of the limit superior and the limit laws
lim sup(xn + yn ) ≤ lim sup xn + lim sup yn .
n→∞
n→∞
n→∞
(b) Explain why lim sup yn = − lim inf (−yn ), and hence
n→∞
n→∞
lim sup(xn − yn ) ≤ lim sup xn − lim inf yn .
n→∞
n→∞
n→∞
For any set A ⊆ R we know that sup A = − inf(−A). Hence, in
Solution:
particular,
sup xk = − inf (−xk )
k≥n
k≥n
for all n ∈ N. Hence by taking limits, using the definition of limit inferior and
limit superior, we get
lim sup xn = lim (sup xk ) = − lim inf (−xk ) = − lim inf (−xn )
n→∞
n→∞ k≥n
n→∞ k≥n
n→∞
as claimed.
(c) Using the previous parts, show that
lim sup(xn + yn ) = lim sup xn + lim sup yn .
n→∞
n→∞
(1)
n→∞
if at least one of the sequences converges.
Solution: Suppose that (yn ) converges (just rename the sequences if (xn ) converges). We know from part (a)
lim sup(xn + yn ) ≤ lim sup xn + lim sup yn = lim xn + lim sup yn .
n→∞
n→∞
n→∞
n→∞
(2)
n→∞
We want to prove the reverse inequality. Applying (b) we get
lim sup xn = lim sup(xn + yn − yn ) ≤ lim sup(xn + yn ) − lim inf (yn ).
n→∞
n→∞
n→∞
n→∞
Hence
lim sup yn + lim inf xn ≤ lim sup(xn + yn ).
n→∞
n→∞
n→∞
Comparing with Combining (2) we get (1) if
lim sup yn = lim inf yn .
n→∞
n→∞
We know that this is the case if (yn ) is convergent.
5
(3)
(d) By giving a counter example, show that strict inequality is possible.
Solution: It is clear from the previous part that both sequences have to diverge. We set xn := (−1)n and yn = (−1)n+1 for all n ∈ N. Then lim sup xn =
n→∞
lim sup yn = 1. On the other hand, xn + yn = (−1)n (1 − 1) = 0 for all n ∈ N, so
n→∞
lim sup(xn + yn ) = 0. Hence,
n→∞
0 = lim sup(xn + yn ) < lim sup xn + lim sup yn = 1 + 1 = 2.
n→∞
n→∞
n→∞
More general examples can be obtained by looking at sequences of the form
(
(
a1 if n is even,
b1 if n is even,
xn =
yn =
a2 if n is odd,
b2 if n is odd,
by choosing a1 , a2 , b1 and b2 appropriately. We have lim supn→∞ xn = max{a1 , a2 },
lim supn→∞ yn = max{b1 , b2 } and lim supn→∞ (xn + yn ) = max{a1 + b1 , a2 + b2 }. By
choosing a1 , a2 , b1 and b2 in R with
max{a1 + b1 , a2 + b2 } < max{a1 , a2 } + max{b1 , b2 },
we obtain a strict inequality in (a). That is exactly what we have done in the first
example: a1 = −1, a2 = 1 and b1 = 1, b2 = −1.
5. Let x ∈ R and consider the sequence given by
x n
xn := 1 +
n
for n ∈ N. Use the arithmetic-geometric mean inequality to solve the following problems.
(a) Suppose that q ∈ N and p > 0. Show that
p q xn ≤
n + x + pq n+q
n+q
for all n ∈ N with n ≥ −x.
(4)
Solution: Note that 1 + x/n ≥ 0 for n ≥ −x. By the arithmetic-geometric mean
inequality with n + q factors we have
x n q n(1 + nx ) + pq n+q n + x + pq n+q
xn p = 1 +
·p ≤
=
n
n+q
n+q
q
for all n ≥ −x.
(b) Show that (xn ) is increasing for all n ∈ N with n ≥ −x.
Solution: If we choose p = q = 1 in (4), then
1
xn = xn 1 ≤
for all n ≥ −x.
n + x + 1 n+1
n+1
6
= 1+
x n+1
= xn+1
n+1
(c) Show that (xn ) is bounded, and therefore it converges.
Solution: We choose p = 1/2 and q such that n + x + q/2 ≤ n + q, that is,
x ≤ q/2. Applying (4) we get
q n+q
n + q n+q
q n+x+ 2
≤ 2q
= 2q
xn ≤ 2
n+q
n+q
for all n ≥ −x.
*(d) For discussion: Can you guess what the limit of (xn ) is? Can you compute it for
certain x ∈ R, and why not for others?
Solution: Formally we can rewrite (xn ) in the form
h
1 nx ix
xn = 1 + n
x
Setting m := n/x it reads
h
1 m ix
,
1+
m
and using the elementary limit (1 + 1/m)m → e as m → ∞ we expect that
xn → ex as n → ∞. Note however, that we needed to assume that m ∈ N for
(1 + 1/m)m → e, and that is not the case for all x. We know that (xn ) converges.
Hence to determine its limit it is sufficient to consider any subsequence (Why?).
Assume that x is rational and positive, that is,
p
x=
q
xn =
for some p, q ∈ N, q 6= 0. Then n/x ∈ N if n is divisible by p, that is n = kp for
k ∈ N. We then have
h
1 kq ip
xqkp = 1 +
→ ep
kq
as k → ∞ by a using a standard limit. If we take the p-th root we get the limit ep/q .
For irrational x we cannot presently compute the limit as pn/x 6∈ N for all p ∈ N.
We could however define ex to be that limit, and then prove all the properties of
the exponential function by starting from x rational, then approximate.
Extra questions for further practice
6. Let (xn ) and (yn ) be bounded sequences in R with xn , yn ≥ 0 for all n ∈ N.
(a) Show that
lim sup(xn yn ) ≤ lim sup xn lim sup yn .
n→∞
n→∞
n→∞
Solution: If ℓ ≥ n, using that xk , yk ≥ 0, we have
xℓ yℓ ≤ sup xk sup yk .
k≥n
k≥n
This means that supk≥n xk supk≥n yk is an upper bound for xℓ yℓ for all ℓ ≥
n. By definition of a supremum (every upper bound is larger or equal than the
supremum)
sup xℓ yℓ ≤ sup xk sup yk
k≥n
ℓ≥n
k≥n
for all n ∈ N. By definition of the limit superior and the limit laws
lim sup(xn yn ) ≤ lim sup xn lim sup yn .
n→∞
n→∞
7
n→∞
(b) By giving a counter example, show that strict inequality is possible.
Solution:
We set xn := 2 + (−1)n and yn = 2 + (−1)n+1 for all n ∈ N.
We then have xn , yn > 0 and lim sup xn = lim sup yn = 3. On the other hand,
n→∞
n→∞
xn yn = (2 + (−1)n )(2 − (−1)n ) = 3 for all n ∈ N, so lim sup(xn yn ) = 3. Hence,
n→∞
3 = lim sup(xn yn ) < lim sup xn lim sup yn = 3 · 3 = 9.
n→∞
n→∞
n→∞
(c) If one of the sequences converge, and the product is not of the form 0 × ∞ or
∞ × 0, show that
lim sup(xn yn ) = lim sup xn lim sup yn .
n→∞
n→∞
n→∞
Solution: Assume that yn → y. If yn → 0 and (xn ) is bounded, then xn yn → 0.
Hence equality holds. Assume therefore that yn → y with y > 0. Applying (a)
lim sup xn lim sup yn = lim sup(xn y)
n→∞
n→∞
n→∞
y
= lim sup (xn yn )
yn
n→∞
y
≤ lim sup(xn yn ) lim sup
(5)
yn
n→∞
n→∞
y
= lim sup(xn yn ) lim
n→∞ yn
n→∞
= lim sup(xn yn )
n→∞
Together with (a) equality follows. The above argument works even if (xn ) is not
bounded from above.
7. *(a) Suppose that (an ) is a sequence in R with an 6= 0 for all n ∈ N. If (an+1 /an )n∈N is
a bounded sequence, prove that
lim inf
n→∞
p
p
|an+1 |
|an+1 |
≤ lim inf n |an | ≤ lim sup n |an | ≤ lim sup
.
n→∞
|an |
|an |
n→∞
n→∞
Solution: We only prove the last inequality. The middle one is obvious from
the definition of the limit superior and inferior, and the first one can be obtained
by reversing the signs. Since (an+1 /an )n∈N is bounded, we have that
s := lim sup
n→∞
|an+1 |
|an |
exists in R. We fix ε > 0. Then by definition of the limit superior there exists
N ≥ 1 (depending on ε) such that
sup
k≥n
|ak+1 |
<s+ε
|ak |
for all n ≥ N . In particular, |ak+1 |/|ak | < s + ε for all k ≥ N . Given n > N we
apply the inequality for each k = N, N + 1, . . . , n − 1 to obtain
|an | |an−1 |
|aN +1 |
|an |
=
···
≤ (s + ε)n−N .
|aN |
|an−1 | |an−2 |
|aN |
8
Therefore
p
n
p
|an | ≤ (s + ε)1−N/n n |aN |
p
for all n > N . We know that n |aN | → 1 as n → ∞ and similarly (s + ε)N/n → 1
as n → ∞. Hence, by the limit laws for the limit superior
p
lim sup n |an | ≤ s + ε.
n→∞
The above argument works for every choice of ε > 0, and hence lim supn→∞
s as claimed.
√
(b) Use part (a) to compute the limit of xn = n n!/n as n → ∞.
Solution: If we set
an :=
then we have
xn =
n!
,
nn
√
n
an
Note that
an+1
(n + 1)! nn n n 1 −n
1
=
=
= 1+
→ .
n+1
an
(n + 1)
n!
n+1
n
e
From part (a) and the squeeze law, we conclude that
lim xn = lim
n→∞
n→∞
9
√
n
1
an+1
= .
n→∞ an
e
an = lim
p
n
|an | ≤