The University of Sydney
School of Mathematics and Statistics
Tutorial 3 (Week 4)
MATH2962: Real and Complex Analysis (Advanced)
Semester 1, 2017
Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/
Lecturer: Florica C. Cı̂rstea
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 2017 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.
n2 − 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.
1. Consider the sequence xn =
(b) Find an = inf k≥n xk and bn = supk≥n xk .
(c) Hence compute lim inf n→∞ xn and lim supn→∞ xn .
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.
1
(b) 1 + (−1)n .
n
n even
n odd
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
X

n even
(−1)n/2
(−1)k
(b) sn =
n+1
2
(a) xn =
n −1
k=0

 2
n odd
2n + 1
4. Let (xn ) and (yn ) be bounded sequences in R.
(a) Let (xn ) and (yn ) be sequences in R. Prove that
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→∞
(c) Using the previous parts, show that
lim sup(xn + yn ) = lim sup xn + lim sup yn .
n→∞
n→∞
n→∞
if at least one of the sequences converges.
(d) By giving a counter example, show that strict inequality is possible.
2
(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
(2)
for all n ∈ N with n ≥ −x.
(b) Show that (xn ) is increasing for all n ∈ N with n ≥ −x.
(c) Show that (xn ) is bounded, and therefore it converges.
*(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?
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→∞
(b) By giving a counter example, show that strict inequality is possible.
(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→∞
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
p
p
|an+1 |
|an+1 |
≤ lim inf n |an | ≤ lim sup n |an | ≤ lim sup
.
n→∞
n→∞
|an |
|an |
n→∞
n→∞
√
(b) Use part (a) to compute the limit of xn = n n!/n as n → ∞.
lim inf
3