Lemma 1 Let {an } be a real sequence such that {an } is bounded above.
(a) lim sup an ≤ l if and only if for any (small) ² > 0, there exists an N = N (²) > 0 such that
n→∞
an < l + ² whenever n ≥ N . Here, l ∈ R.
(b) lim sup an ≥ l if and only if for any (small) ² > 0, there exists a subsequence {ank } of
n→∞
{an } such that ank > l − ² for nk À 1. Here, l ∈ R.
Denote by A the set of all subsequential limits of {an }. Since {an } is bounded above, it
follows that an ≤ M for some constant M > 0.
Proof of (a).
(⇒) We argue by contradiction and suppose that there exist a constant ²0 > 0 satisfying the
following property:
For any N > 0, there exists an n ≥ N such that an ≥ l + ²0 .
We claim that there exists a subsequence {ank } of {an } such that ank ≥ l + ²0 . Indeed,
there exists an n1 ≥ 1 such that an1 ≥ l + ²0 . Moreover, there exists an n2 ≥ n1 + 1 such that
an2 ≥ l + ²0 . Repeating this argument, we can choose an nk ≥ nk−1 + 1 such that ank ≥ l + ²0 .
Our claim is proved.
We claim that A contains an element x0 ∈ [l + ²0 , M ]. Indeed, {ank } ⊂ [l + ²0 , M ] and the
interval [l+²0 , M ] is compact. Then {ank } has a subsequence {bn } which converges to an element
in [l + ²0 , M ]. Let x0 = lim bn . Since l + ²0 ≤ ank ≤ M , it follows that l + ²0 ≤ x0 ≤ M .
n→∞
Notice that {bn } is a subsequence of {an }, and consequently, x0 is a subsequential limit of {an }.
Namely x0 ∈ A ∩ [l + ²0 , M ]. Our claim is proved.
Therefore, we conclude that l + ²0 ≤ x0 ≤ sup A = lim sup an ≤ l, which leads to a contradicn→∞
tion.
////
(⇐) We claim that sup A ≤ l + ² for any ² > 0.
Indeed, let ² > 0 be given. Choose any element x ∈ A. Since x is a subsequential limit of
{an }, there exists a subsequence {ank } of {an } such that lim ank = x. Then there exists an
nk →∞
N such that l − ² < ank < l + ² for nk ≥ N . Consequently x = lim ank ≤ l + ².
nk →∞
We have proved that x ≤ l + ² for any x ∈ A. Then it follows that sup A ≤ l + ². Our claim
is proved.
Since the choice of ² > 0 is arbitrary, we conclude that sup A ≤ l.
¤
Proof of (b).
(⇒) Note that sup A = lim sup an . Given ² > 0, there exists an element y ∈ A such that
n→∞
y > sup A − ² ≥ l − ². Since y is a subsequential limit of {an }, there exists a subsequence {ank }
such that lim ank = y. Since y > l − ², it follows that ank > l − ² for nk À 1.
////
nk →∞
(⇐) Let {ank } be a subsequence of {an } such that ank > l − ² for nk À 1. Since an ≤ M , it
follows that l − ² ≤ ank ≤ M for nk À 1. In other words, {ank } ⊂ [l − ², M ] for nk À 1. Since
[l − ², M ] is compact, {ank } has a subsequence {bn } such that {bn } converges to an element of
[l − ², M ].
1
Let x0 = lim bn . Note that x0 ∈ [l − ², M ] and x0 is a subsequential limit of {an }, in other
n→∞
words, x0 ∈ A. Therefore it follows that
lim sup an = sup A ≥ x0 ≥ l − ².
n→∞
Since ² > 0 is arbitrary, we conclude that lim sup an ≥ l.
¤
n→∞
Exercieses 6.1
(You may skip #2, #4 and all the problems related to lim inf except #15 and #16. Lemma 1,
n→∞
#7-(a), #9-(c), #14-(a) and #16 are strongly recommended. See Remark 3 as well.)
1. Let an = αn + iβn where αn , βn ∈ R. Show that
∞
X
n=1
|αn | and
∞
X
∞
X
|an | converges if and only if both
n=1
|βn | converge.
n=1
Solution. Note that |αn |, |βn | ≤ |an | ≤ |αn | + |βn | for all n. For the sake of convenience,
n
∞
n
X
X
X
we let sn =
|αk |, tn =
|βk | and un =
|ak |. Then {sn }, {tn } and {un } are
k=1
k=1
k=1
monotonically increasing, and moreover, sn , tn ≤ un ≤ sn + tn for all n.
(⇒) Since {un } converges, {un } is bounded. Since sn , tn ≤ un , it follows that {sn } and
{tn } are both bounded above. Consequently both {sn } and {tn } are convergent.
(⇐) Since both {sn } and {tn } converge, they are bounded. Since un ≤ sn + tn , {un } is
bounded above. Consequently {un } converges.
¤
P∞
3. Suppose an > 0 for every n. Show that n=1 an diverges ⇔ for any integers M and N ,
PN +p
there exists an integer p such that n=N an > M .
Solution. Let sn =
n
X
ak for simplicity. Since sn is monotonically increasing, {sn } diverges
k=1
if and only if sn → ∞.
(⇒) Let M, N ∈ N be given. Since sn → ∞, it follows that sn − sN → ∞. Consequently
N
+p
X
there exists an integer p such that sN +p − sN > M . In other words,
an > M .
n=N
(⇐) We argue by contradiction, and suppose that {sn } converges. Then {sn } is bounded
above, and there exists a constant M > 0 such that sn ≤ M for all n. Consequently, for
N
+p
X
any integers N and p,
an = sN +p −sN −1 ≤ sN +p ≤ M , which yields a contradiction. ¤
n=N
5. Let A be the set of subsequential limits of a complex sequence {an }. Show that A is closed.
Proof. We need to prove that A ⊂ A. Choose any x ∈ A. We need to find a subsequence
{ank } of {an } such that ank → x.
Let ² > 0 be given. Then there exists an element y ∈ A such that |y −x| < ²/2. Since y ∈ A,
2
we can choose a subsequence {amj } such that amj → y. Then there is an N = N (²) > 0
such that |amj − y| < ²/2 for mj ≥ N . Consequently
|amj − x| ≤ |amj − y| + |y − x| < ²
for mj ≥ N.
We have proved that for any ² > 0 there is an N 0 = N 0 (²) such that |aN 0 − x| < ².
Therefore, for each k ∈ N, we can choose an nk ∈ N such that |ank − x| < 1/k. It is obvious
that ank → x as k → ∞, and {ank } is the desired subsequence.
¤
6. (left as an exercise: homework) Note that lim an = 0 if and only if
n→∞
lim |an | = 0.
n→∞
7. (a) If an ≥ bn for every n, show that
lim sup an ≥ lim sup bn .
n→∞
n→∞
Proof. We consider the following three cases.
Case 1. lim sup an = l ∈ R:
n→∞
For any ² > 0, there exists an N = N (²) > 0 such that
an ≤ l + ² for n ≥ N (see Lemma 1). Since bn ≤ an , it follows that bn ≤ l + ² for n ≥ N .
Then Lemma 1-(a) implies that lim sup bn ≤ l.
n→∞
Case 2. If lim sup an = ∞ then there is nothing to prove.
n→∞
Case 3. If lim sup an = −∞ then an → −∞. Since an ≥ bn it follows that bn → −∞, and
n→∞
consequently, lim sup bn = −∞.
¤
∗ Remark 1. The conclusion also holds true if we assume that an ≥ bn for n À 1.
♣
n→∞
∗ Remark 2. Recall that lim sup bn = l = lim inf bn if and only if
n→∞
n→∞
lim bn = l.
n→∞
8. (a) Let {an } and {bn } be positive, bounded sequences. Show that
³
´³
´
lim sup(an bn ) ≤ lim sup an lim sup bn .
n→∞
n→∞
n→∞
Proof. Let A be the set of subsequential limits of {an bn }. Since {an } and {bn } are bounded,
it follows that {an bn } is bounded, and consequently A is also bounded. For simplicity, we
let l = lim sup an and m = lim sup bn . Clearly l, m ≥ 0.
n→∞
n→∞
Let ² > 0 be given. By Lemma 1, there exists an N = N (²) > 0 such that an < l + ²
and bn < m + ² for all n ≥ N . Then it follows that an bn < (l + ²)(m + ²) for n ≥ N .
Consequently, #7-(a) (Remark 1) implies that
lim sup(an bn ) ≤ lim sup[(l + ²)(m + ²)] = (l + ²)(m + ²).
n→∞
n→∞
Since ² > 0 is arbitrary, we conclude that lim sup(an bn ) ≤ lm.
n→∞
3
¤
9. (a) (left as an exercise: homework)
(c) (Modified problem) Suppose that {an } is real and lim an = a ≥ 0. For any real bounded
n→∞
sequence {bn }, show that
³
lim sup(an bn ) =
¶
´µ
lim an
lim sup bn
n→∞
n→∞
n→∞
Solution. We consider two cases:
Case 1: a > 0
Let l = lim sup bn for simplicity. Let 0 < ² < a be given. Lemma 1 implies that
n→∞
(
there is an N1 = N1 (²) > 0 such that bn < l + ² for n ≥ N1 ,
there exists a subsequence {bnk } such that bnk > l − ² for nk À 1.
Moreover, there is an N2 = N2 (²) > 0 such that |an − a| < ² for n ≥ N2 . Let N =
max{N1 , N2 }. If n ≥ N then an > a − ² > 0, and consequently,
(
an bn < (a + ²)(l + ²) = al + ²(a + l + ²) for n ≥ N,
(1)
ank bnk > (a − ²)(l − ²) = al − ²(a + l − ²) for nk ≥ N.
Then it follows from Lemma 1 that
al ≤ lim sup(an bn ) ≤ al.
n→∞
³
Therefore, lim sup(an bn ) = al =
n→∞
lim an
n→∞
¶
´µ
lim sup bn .
n→∞
Case 2: a = 0
Since {bn } is bounded, there exists a constant M > 0 such that |bn | ≤ M for all n. Then
it follows that 0 ≤ |an bn | = |an ||bn | ≤ M |an | → 0 as n → ∞. Consequently an bn → 0 as
n → ∞. Since −M ≤ lim sup bn ≤ M , we conclude that
n→∞
³
lim sup(an bn ) = lim (an bn ) = 0 =
n→∞
n→∞
¶
´µ
lim an
lim sup bn .
n→∞
n→∞
¤
12. Show that lim sup an = L, L finite, if and only if the following conditions hold: For any ² > 0,
n→∞
(i) There exists an N = N (²) > 0 such that an < L + ² for all n ≥ N ;
(ii) There exists a subsequence {ank } such that ank > L − ². ({nk } is infinite.)
Solution. #12 is a Corollary of Lemma 1.
14. Let {an } be a complex sequence.
∞
¯a
¯
X
¯ n+1 ¯
(a) If lim sup ¯
an converges absolutely.
¯ = L < 1 show that
an
n→∞
n=1
4
¤
Solution. Fix a constant
¯a
¯ r > 0 such that L < r < 1. Lemma 1 implies that there exists an
¯ n+1 ¯
N > 0 such that ¯
¯ < r for n ≥ N (² = r − N ). In other words, |an+1 | < r|an | for
an
n ≥ N . Therefore, if n > N then
|an | < r|an−1 | < r2 |an−2 | < · · · < rn−N |aN |.
Since 0 < r < 1, it follows that
that
∞
X
∞
X
rn−N |aN | converges. Then the comparison test shows
n=1
|an | also converges. Therefore
n=1
∞
X
an converges absolutely.
¤
n=1
¯a
¯
P
¯ n+1 ¯
Remark: If lim sup ¯
an .
¯ > 1 then we have no information on the convergence of
an
n→∞
For example, if we let a2n−1 = 1/2n and a2n = 1/3n then
lim sup
n→∞
|an+1 |
= ∞ > 1.
|an |
See #16. However,
∞
X
an =
n=1
1 1
1
1
1
1
+ +
+ 2 + 3 + 3 + ···
2 3 22
3
2
3
converges.
♣
15. Suppose |an | > 0 for every n. Show that
lim inf
n→∞
|an+1 |
|an+1 |
≤ lim inf |an |1/n ≤ lim sup |an |1/n ≤ lim sup
.
n→∞
|an |
|an |
n→∞
n→∞
Proof. We prove the third inequality only. If |an+1 |/|an | is not bounded above, namely if
|an+1 |
lim sup
= ∞ then there is nothing to prove.
|an |
n→∞
|an+1 |
Suppose that |an+1 |/|an | is bounded above, namely, lim sup
= l for some l ∈ [0, ∞).
|an |
n→∞
Let ² > 0 be given. There is an N = N (²) > 0 such that |an+1 |/|an | < l + ² for n ≥ N .
Then, for any n > N ,
|an | < (l + ²)|an−1 | < (l + ²)2 |an−2 | < · · · < (l + ²)n−N |aN |,
n ≥ N.
Consequently,
N
|an |1/n < (l + ²)1− n |aN |1/n
for any n ≥ N . Since (l + ²)1−N/n |aN |1/n converges to l + ² as n → ∞, #7-(a) (Remark 1)
implies that
N
N
lim sup |an |1/n ≤ lim sup[(l + ²)1− n |aN |1/n ] = lim [(l + ²)1− n |aN |1/n ] = l + ².
n→∞
n→∞
n→∞
Since ² > 0 is arbitrary, we conclude that lim sup |an |1/n ≤ l.
n→∞
5
¤
∗ Remark 3. If lim |an+1 |/|an | exists, #15 implies that lim |an |1/n also exists. Moreover,
n→∞
n→∞
lim |an |1/n = lim
n→∞
For example, consider a sequence an =
n→∞
|an+1 |
.
|an |
2n n!
. Since
nn
an+1
2n+1 (n + 1)! nn
2nn
2
¢n ,
=
· n =
=¡
n+1
n
an
(n + 1)
2 n!
(n + 1)
1 + n1
|an+1 |
2
= . Therefore lim sup |an |1/n = lim |an |1/n = 2/e.
n→∞
|an |
e
n→∞
This fact is useful in finding a radius of convergence (Section 6.3).
it follows that lim
n→∞
16. Define a sequence {ak } by a2k−1 =
lim inf
n→∞
an+1
= 0,
an
lim sup
n→∞
♣
1
1
and a2k = k for every k. Show that
k
2
3
an+1
= ∞,
an
lim inf (an )1/n = 0,
n→∞
lim sup(an )1/n = 0.
n→∞
Solution. Note that




2k
a2k
= k,
n = 2k − 1,
an+1
a2k−1
3
=
3k
1 ³ 3 ´k
a

an

 2k+1 = k+1 =
, n = 2k.
a2k
2
2 2
Consequently an+1 /an → 0 if n = 2k − 1 → ∞, and an+1 /an → ∞ if n = 2k → ∞.
Therefore
an+1
an+1
lim inf
= 0,
lim sup
= ∞.
n→∞
an
an
n→∞
We also note that

³ ´k/(2k−1)

 (a2k−1 )1/(2k−1) = 1
, n = 2k − 1,
(an )1/n =
³ 1 ´1/22

 (a2k )1/(2k) =
,
n = 2k.
3
√
√
Consequently (an )1/n → 1/ 2 if n = 2k − 1 → ∞, and (an )1/n → 1/ 3 if n = 2k → ∞.
Therefore
1
1
lim inf (an )1/n = √ ,
lim sup(an )1/n = √ .
n→∞
n→∞
3
2
¤
6