c University of Bristol 2014
Further Topics in Analysis: Solutions 6
1. Let (an )n∈N be a sequence in (0, ∞). Set
l := lim sup a1/n
n .
n→∞
P
(i) Show that if 0 6 l < 1, then ∞
n=1 an converges.
P∞
(ii) Show that if l > 1, then n=1 an diverges.
(iii) Find λ ∈ [0, ∞) for which
∞
X
n
2n(−1) λn
n=1
converges respectively diverges.
Hint: For (i) and (ii), adapt the proof of the root test.
Solution. (i) Assume that 0 6 l < 1. Choose r ∈ R with l < r < 1. Put
1/k
(n ∈ N+ ).
bn := sup ak
k>n
Then (bn )n∈N+ decreases to l. Thus
(∃ N ∈ N+ )(∀ n ∈ N+ )[ (n > N ) ⇒ (bn < r) ].
Then for any N+ 3 k > N ,
1/k
ak
1/j
6 sup aj
= bN < r
j>N
and
ak < rk for all k > N.
P∞
So
P
∞
ak converges by comparision with the geometric series
a
converges.
k=1 k
k=N
P∞
k=N
rk . It then follows that
(ii) Assume that l > 1. Choose r ∈ R with l > r > 1. Now l is an accumulation point of
1/n
1/m(k)
(an )n∈N+ . So there exists a subsequence (am(k) )k∈N+ such that
1/m(k)
am(k)
→ l as k → ∞.
In particular,
1/m(k)
(∃ K ∈ N+ )(∀ k ∈ N+ )[(k > K) ⇒ (am(k)
> r) ]
or
am(k) > rm(k) > 1 for all k > K.
It cannot be the case therefore that limn→∞ an = 0. Consequently,
(iii) We have
n
P∞
l := lim sup a1/n
= lim sup 2(−1) λ = 2λ.
n
n→∞
n→∞
So the series converges for 0 6 λ < 1/2 and diverges for λ > 1/2.
n=1
an diverges.
2. Let (an )n∈N+ and (bn )n∈N+ be bounded sequences in R.
(a) Show that
lim inf { an − bn } > lim inf an − lim sup bn .
n→∞
n→∞
n→∞
It may be helpful to use the formula given in lectures.
(b) Formulate a counterpart to this result for the limit superior.
Solution.
(a) By the formula given in lectures,
lim inf { an − bn } = lim inf { ak − bk } .
n→∞
n→∞ k>n
Set
αn := inf ak and βn := sup bk .
k>n
k>n
For any k > n, ak > αn and bk 6 βn ; so
ak − bk > αn − βn ;
hence,
inf { ak − bk } > αn − βn = inf ak − sup bk .
k>n
k>n
k>n
Therefore,
lim inf { an − bn }
n→∞
>
=
=
inf ak − sup bk
lim
n→∞
k>n
k>n
lim inf ak − lim sup bk
n→∞ k>n
n→∞ k>n
lim inf an − lim sup bn .
n→∞
n→∞
(b)
lim sup { an − bn } 6 lim sup an − lim inf bn .
n→∞
n→∞
n→∞
3. Let (an )n∈N+ and (bn )n∈N+ be bounded sequences in [0, +∞).
(a) Show that
lim sup an bn 6 lim sup an · lim sup bn .
n→∞
n→∞
n→∞
(b) Demonstrate that the inequality in (a) need not hold for bounded sequences in R.
Solution.
(a) We have
lim sup an bn
=
n→∞
lim sup ak · bk
n→∞ k>n
6
=
=
lim
n→∞
k>n
k>n
lim sup ak · lim sup bk
n→∞ k>n
n→∞ k>n
lim sup an · lim sup bn .
n→∞
2
sup ak · sup bk
n→∞
(b) Suppose bn = −1 for all n ∈ N+ . The inequality then reads
lim sup { −an } 6 − lim sup an .
n→∞
n→∞
Now it is straightforward to see that
lim sup { −an } = − lim inf an ,
n→∞
n→∞
so the above inequality becomes
− lim inf an 6 − lim sup an
n→∞
n→∞
or
lim inf an > lim sup an .
n→∞
n→∞
This inequality holds if and only if (an ) converges. It does not hold in general. Take, for
example, an = (−1)n .
4. Describe the set of all accumulation points and find the limit superior and limit inferior
of the following sequences.
√
√
(a) an = n + 1 + (−1)n n − 1;
√
(b) an =
(c) an =
√
n+1+(−1)n n−1
;
n
√
√
n+1+(−1)n n−1
√
;
n
(d) (an )n∈N = (0, 1, 0, −1, 0, 1, 0, −1, 0, 1, 0, −1, . . . ).
Solution.
(a) The sequence (an )n∈N is unbounded but has a convergent subsequence
√
a2k−1 =
2k −
√
2k − 2 = √
2k +
2
√
2k − 2
→0
as k → ∞. The set of all accumulation points of (an )n∈N is {0} (you should convince yourself
that there are no other accumulation points).
In the lecture notes we do not define the limit superior and the limit inferior of unbounded
sequences, so we would be justified in saying that the limit superior and limit inferior in this
example do not exist. Note, however, that for the limit inferior to exist we only require the
sequence to be bounded below. Since in this example an > 0 for all n ∈ N, and 0 is indeed an
accumulation point, we conclude that lim inf n→∞ an = 0. The limit superior of this sequence,
on the other hand, does not exist as the even terms form an unbounded subsequence. (I will
have explained this in the lecture.)
(b) The sequence (an )n∈N converges to 0 as n → ∞. Indeed, we see that
r
√
√
√
n+1+ n−1
n+1
1
1
|an | ≤
≤2
=2
+
→0
n
n
n n2
as k → ∞. Thus the set of all accumulation points of (an )n∈N is {0}.
If the sequence converges then the limit superior and limit inferior coincide with the limit (see
Lemma 7.6), so
lim sup an = lim inf an = lim an = 0.
n→∞
n→∞
3
n→∞
(c) The sequence (an )n∈N is bounded and has convergent subsequences
r
r
√
√
2k − 2k − 2
1
1
√
a2k−1 =
= 1+
− 1−
→ 0,
2k
−
1
2k
+1
2k − 1
r
r
√
√
1
1
2k + 1 + 2k − 1
√
= 1+
a2k =
+ 1−
→ 2,
2k
2k
2k
as k → ∞, and indeed 0 and 2 are the only accumulation points.
In this case
lim sup an = sup{0, 2} = 2,
lim inf an = inf{0, 2} = 0.
n→∞
n→∞
(d) It is easy to see that the set of all accumulation points of (an )n∈N is {−1, 0, 1}.
In this case
lim sup an = sup{−1, 0, 1} = 1,
n→∞
lim inf an = inf{−1, 0, 1} = −1.
n→∞
5. Let (an )n∈N be a bounded sequence of real numbers. Which of the following must be
true? Give counterexamples for those that may be false.
(a) lim supn→∞ an = lim inf n→∞ an .
(b) (an )n∈N has a convergent subsequence tending to lim supn→∞ an .
(c) If (bm )m∈N is a convergent subsequence of (an )n∈N , then either
bm → lim sup an
n→∞
or bm → lim inf an
n→∞
as m → ∞.
(d) If (bm )m∈N is a convergent subsequence of (an )n∈N , with limm→∞ bm = b then
lim inf an 6 b 6 lim sup an .
n→∞
n→∞
(e) For every real number b with
lim inf an 6 b 6 lim sup an ,
n→∞
n→∞
there is a convergent subsequence (bm )m∈N of (an )n∈N such that limm→∞ bm = b.
Solution.
(a) FALSE. E.g., an = (−1)n .
(b) TRUE. By Theorem 9.5 we know that if (an )n∈N is a bounded sequence then lim supn→∞ an
and lim inf n→∞ an are accumulation points of (an )n∈N .
(c) FALSE. E.g., if

 −1
0
an =

1
if n ≡ −1 (mod 3)
if n ≡ 0 (mod 3)
if n ≡ 1 (mod 3),
then (a3n )n∈N is a subsequence tending to 0, whereas
lim sup an = 1 and
n→∞
4
lim inf an = −1.
n→∞
(d) TRUE. This follows from the definition of the limit superior and limit inferior.
(e) FALSE. E.g., if an = (−1)n , then
lim inf an = −1
n→∞
and
lim sup an = 1,
n→∞
but there is no subsequence tending to 0 (or to any other real number b with −1 < b < 1).
6. Determine whether or not the following sequences (an )n∈N are Cauchy sequences, and
justify your answer directly from the definition.
(a) an =
n
n+1 ;
(b) an =
2n +1
2n ;
(−1)n (2
(c) an =
+
1
).
n2
Solution. (a) Fix ε > 0. Choose N ∈ N such that N > 1/ε. Then for any m, n ≥ N we obtain
m
1
1
n 1
1
1
≤ max
1
−
−
=
−
1
−
,
<
< ε.
|an −am | = m + 1 n + 1 m+1
n+1 m+1 n+1
N
This means that (an )n∈N is a Cauchy sequence.
(b) Fix ε > 0 and take N ∈ N such that 2N > 1/ε. Let m, n ≥ N . Then
m
2 + 1 2n + 1 1
1 1 1
1
=
−
≤
max
,
≤ N < ε.
|am − an | = m −
2
2n 2m
2n 2m 2n
2
This means that (an )n∈N is a Cauchy sequence.
(c) We need to verify the negation of the definition of a Cauchy sequence, i.e.
(∃ε > 0)(∀N ∈ N)(∃m, n ∈ N)[(m ≥ N ) ∧ (n ≥ N ) ∧ (|am − an | ≥ ε)].
Fix ε = 4 (any other positive ε < 4 will work too). For any N ∈ N set n := N and m := N + 1.
Then
|an − am | = |aN − aN +1 | > 4 = ε.
Thus (an )n∈N is not a Cauchy sequence.
7. (a) Let (an )n∈N be a sequence such that
(∀n ∈ N)(|an+1 − an | 6
1
).
2n
Prove that (an )n∈N is a Cauchy sequence.
(b) Let (an )n∈N be the sequence defined by a1 = 1, a2 = 1/2 and
1
an+2 = (an+1 + an ) for all n > 1.
2
Prove that (an )n∈N converges.
5
Solution. (a) Fix ε > 0. Choose N ∈ N such that N > 1/ε + 1. Then by using the triangle
inequality and the formula
∞ k
∞
X
1 X 1
1
1
1
= N
= N
k
2
2
2
2 1−
k=0
k=N
1
2
=
1
2N −1
for any m, n ≥ N we obtain
|an − am |
≤
≤ |an − an+1 | + |an+1 − an+2 | + · · · + |am−1 − am |
≤
m−1
X
k=n
∞
X
1
1
1
1
≤
= N −1 6
< ε.
k
k
2
2
2
N −1
k=N
(b) It is enough, by Cauchy’s theorem, to prove that (an )n∈N is a Cauchy sequence. We will prove
by induction that |an+1 − an | = 21n . For n = 1 this is clear. For n > 2 we have |an+1 − an | =
1
| 21 (an + an−1 ) − an | = 12 |an−1 − an | which equals 12 2n−1
= 21n by induction. By (a) we are done.
6