5
Sequences
Let (X, d) be a metric space. A sequence in X is a function from N to X.
If ϕ : N → X is a sequence we write
(xn ), (xn )n∈N
or
(x0 , x1 , . . .)
for ϕ, where xn = ϕ(n) is the nth term of the sequence ϕ = (x0 , x1 , . . .).
Sequences in R are called real sequences.
Definition 5.1. A sequence (xn ) in X converges to x if for every neighborhood U of x there exists N ∈ N such that
xn ∈ U
for all n ≥ N .
In this case we write
xn → x
or
lim xn = x
n→∞
or
lim xn = x.
n
A sequence (xn ) which does not converge is called divergent.
Proposition 5.2. The following are equivalent.
(i) limn xn = x.
(ii) For every ε > 0, there is N ∈ N such that xn ∈ B(x, ε) for all n ≥ N .
(iii) For every ε > 0, there is N ∈ N such that d(xn , x) < ε for all n ≥ N
Proposition 5.3. (i) (Uniqueness of the limit) A sequence cannot have
more than one limit.
(ii) (Boundedness) If (xn ) converges to x, then there is r > 0 such that
xn ∈ Br (x).
Proof of (ii). Assume that (xn ) converges to x. By (iii) of the previous
proposition with ε = 1, there is N so that xn ∈ B1 (x) for all n ≥ N . Now
take r be any number bigger or equal to d(x1 , x), . . . , d(xN −1 , x) and 1. Then
xn ∈ Br (x) for all n, as claimed.
Definition 5.4. Let ϕ = (xn ) be a sequence in X and let φ : N → N be a
strictly increasing function. Then ϕ ◦ φ : N → X is called a subsequence of
ϕ = (xn ).
We write (xnk ) for the subsequece ϕ ◦ φ where φ(k) = nk . Since ψ is
strictly increasing n1 < n2 < n3 < . . ..
20
Proposition 5.5. If (xn ) is a convergent sequence with limit x, then every
subsequence (xnk ) of (xn ) converges to x.
Proof. Let (xnk ) be a subsequence of (xn ). Take a neighborhood U of x.
Since lim xn = x, there is N ∈ N such that xn ∈ U for all n ≥ N . From the
definition of a subsequence, nk ≥ k for all k ∈ N. So, nk ≥ N for all k ≥ N .
Thus xnk ∈ U for all k ≥ N . Hence (xnk ) converges to x as claimed.
Definition 5.6. A sequence (xn ) in a metric space (X, d) is said to be
Cauchy if for every ε > 0, there exists N such that
d(xn , xm ) ≤ ε for all n, m ≥ N.
Proposition 5.7. Let (X, d) be a metric space.
(i) Every Cauchy sequence is bounded.
(ii) Every convergent sequence in X is Cauchy.
Proof of (ii). Assume that (xn ) converges to x. Then, given ε > 0, there
is N such that
d(xn , x) < ε/2 for all n ≥ N..
Hence
d(xn , xm ) ≤ d(xn , x) + d(x, xm ) < ε/2 + ε/2 = ε
for all n, m ≥ N . So, (xn ) is a Cauchy sequence as claimed.
5.1
Real sequences
Proposition 5.8. Let (xn ) and (yn ) be convergent sequences in R and let
α ∈ R. Then
(i) The sequence (xn + yn ) converges and lim(xn + yn ) = lim xn + lim yn .
(ii) The sequence (αxn ) converges and lim(αxn ) = α lim xn .
Proof of (i). Take ε > 0. Since (xn ) and (yn ) converge to x and y, respectively, there are N1 and N2 such that
|xn − x| < ε/2
for n ≥ N1
and
|yn − y| < ε/2
for n ≥ N2 .
Consequently,
|(xn + yn ) − (x + y)| = |(xn − x) + (yn − y)| ≤ |xn − x| + |yn − y|
≤ ε/2 + ε/2 = ε
for all n ≥ N = max{N1 , N2 }. Hence the sequence (xn + yn ) converges and
lim(xn + yn ) = lim xn + lim yn , as claimed.
21
Note that above proposition holds true if instead of real sequences one
considers sequences in a normed space (X, k·k). We call a real sequence (xn )
a null sequence if it converges to 0. That is, for every ε > 0, there is N ∈ N
such that |xn | < ε for all n ≥ N .
Proposition 5.9. Let (xn ) and (yn ) be sequences in R.
(i) If (xn ) is a null sequence and (yn ) is a bounded sequence, then the
sequence (xn yn ) is a null sequence.
(ii) If (xn ) and (yn ) are convergent sequences, then the sequence (xn yn )
converges and lim(xn yn ) = lim xn lim yn .
(iii) If (xn ) converges to x and x 6= 0, then almost all terms of (xn ) are
nonzero and the sequence (1/xn ) converges to 1/x.
Proof of (iI). Suppose that lim xn = x and lim yn = y. By (ii) of Proposition 5.3, |xn | ≤ A and |yn | ≤ B for all n. Moreover, given ε > 0, there are
N1 and N2 so that
|xn − x| < ε/2(A+|y|) for n ≥ N1
and
|yn − y| < ε/2(A+|y|) for n ≥ N2 .
Consequently,
|(xn yn ) − (xy)| = |xn (yn − y) + y(xn − x)| ≤ |xn | |yn − y| + |y| |xn − x|
ε
ε
+ |y|
=ε
≤ A |yn − y| + |y| |xn − x| ≤ A
2(A + |y|)
2(A + |y|)
for all n ≥ N = max{N1 , N2 }. Hence the sequence (xn yn ) converges and
lim(xn yn ) = lim xn lim yn , as claimed.
Proposition 5.10. (i) Let (xn ) and (yn ) be convergent sequences in R
and xn ≤ yn for infinitely many n. Then lim xn ≤ lim yn .
(ii) Let (xn ) and (yn ) and (zn ) be sequences in R and xn ≤ yn ≤ zn for
almost all n. If lim xn = lim zn = a, then (yn ) converges to a.
Proof of (ii). Take ε > 0. Then here are N1 and N2 so that
|xn − a| < ε for n ≥ N1
and
|zn − a| < ε
for n ≥ N2 .
Hence
a − ε < xn < a + ε
for n ≥ N1
and a − ε < zn < a + ε for n ≥ N2
22
and since xn ≤ yn ≤ zn ,
a − ε < xn ≤ yn ≤ zn < a + ε
for all n ≥ N = max{N1 , N2 }. This means that |yn − a| < ε for all n ≥
N.
Definition 5.11. A sequence (xn ) in R is increasing, if xn ≤ xn+1 for all
n ∈ N, and decreasing if xn+1 ≤ xn for all n ∈ N. If (xn ) is either increasing
or decreasing, then it is called monotone.
Proposition 5.12. Every increasing (or decreasing) bounded sequence (xn )
in R converges,and
xn ր sup{xn | n ∈ N}
( or xn ց inf{xn | n ∈ N}).
Proof. Suppose that (xn ) is increasing and Set a = sup{xn | n ∈ N} ∈ R. By
the characterization of the supremum,
(a) xn ≤ a for all n.
(b) given ε, there is N such that a − ε < xN .
Since (xn ) is increasing,
a − ε < xN ≤ xn ≤ a < a + ε
for all n ≥ N . Hence (xn ) converges to x as claimed.
Example 5.13. Let a > 0 and α > 0. Define a sequence (xn ) by setting x0 = α
and
a
xn +
.
(1)
xn
√
We claim that the sequence (xn√
) converges to a. To see this we shall show
that (xn ) is bounded
below by a and it is decreasing. Consider the function
one finds that f ′ (x) = 12 1 − xa2 < 0
f (x) = 12 x√+ xa for x > 0. Differentiating
√
′
for x√∈ (0, a) and f√
(x) > √
0 for x ∈ ( a,
√ ∞). Hence f takes its minimum at
a = a and f (x) ≥ f ( a) = a. So, xn ≥ a for all n ≥ 1. Next
1
a
a
1
x2 − a
≥0
xn − xn+1 = xn −
xn +
xn −
=
= n
2
xn
2
xn
2xn
xn+1 =
1
2
since x2n ≥ a. This means that xn ≥ xn+1 for all n ≥ 1 so that √
(xn ) is decreasing.
Then the above proposition implies that xn → x where x ≥ a > 0 Then also
xn+1 → x. So, taking limits in (1), we get
1
a
x=
x+
2
x
√
and solving for x one finds x = a.
23
Proposition 5.14. Every sequence contains a monotone subsequence.
Proof. Consider a sequence (xn ). We call the m-th term xm a peak if
xm ≥ xn for all n ≥ m. There are two cases to consider.
Case 1.
There are infinitely many peaks. List them according the increasing subscripts: xn1 , xn2 , . . .. Then, since each of them is a peak, we
have
xn 1 ≥ xn 2 ≥ xn 3 ≥ . . . .
Hence (xnk ) is a decreasing subsequence of (xn ).
Case 2. There are finitely many peaks. List them according to the increasing xm1 , xm2 , . . . , xml . Then set n1 = ml + 1. The n1 terms xn1 is not a
peak and so there is an index n2 > n1 such that xn1 < xn2 . The term xn2 is
also not a peak and there is n3 > n2 such that xn2 < xn3 . Continuing this
way we obtain a strictly increasing subsequence (xnk ) of (xn ).
5.1.1
Infinite limits
Definition 5.15. A sequence of real numbers (xn ) converges to ∞, in symbols xn → ∞ or lim xn = ∞, if for every M > 0 there exists N such that
xn ≥ M
for all n ≥ N .
Similarly, (xn ) converges to −∞, in symbols xn → −∞ or lim xn = −∞ if
for every M < 0 , there is there exists N such that
xn ≤ M
5.1.2
for all n ≥ N .
Some Special Sequences
(1) Let a ∈ R. Then
an → 0 if |a| < 1
an diverges
Suppose that |a| < 1. Then |a| =
1
1+h
if |a| > 1
for some h > 0. By the binomial theorem,
(1 + h)n =
n X
n k
h ≥ nh
k
k=0
so that
n
0 ≤ |an | = |a| =
24
1
1
≤
.
(1 + h)n
nh
Since 1/nh → 0, an → 0 by the comparison test. Assume that |a| > 1. Then
|a| = 1 + h for some h and then
|an | = |an | = (1 + h)n ≥ nh,
showing that (an ) is unbounded.
(2) Let k ∈ N and let a ∈ R be such that |a| > 1. Then
nk
= 0.
n→∞ an
lim
To see this write a = 1 + h where h > 0 and apply the binomial theorem,
n X
n m
n
n(n − 1) · · · (n − k) k+1
n
n
h
.
a = (1 + h) =
h ≥
hk+1 ≥
(k + 1)!
m
k
+
1
m=0
So,
0≤
nk
(k + 1)!nk
≤ k+1
k
a
h
n(n − 1) · · · (n − k)
and the claimed follows using the comparison since the right hand side converges
to 0 as n → ∞.
(3) For all a ∈ R,
an
= 0.
lim
n→∞ n!
√
(4) limn→∞ n n =
√1.
√
To see this write n n = 1 + an with an ≥ 0 (since it is easy to see that n n ≥ 1. So,
(1 + an )n = n.
By the binomial theorem,
n
n 2
n(n − 1) 2
n 2
an .
n = (1 + an )n = 1 +
an +
an + . . . ≥ 1 +
an = 1 +
2
1
2
2
From this we get
0 ≤ a2n ≤
2
,
n
i.e., 0 ≤
p
√
n
n − 1 = an ≤ 2/n
√
which by comparison implies
that n n → 1.
√
n
(5) For all a > 0, limn→∞ a = 1.
If a = 1, then there is nothing to prove. If a > 1, then
√
√
1na≤ nn
√
which implies
that ( n a) converges
to 1. So, assume that 0 < a < 1, then 1/a > 1
p
p
√
so that n 1/a → 1. Since n a = 1/ n 1/a, the claim follows.
25
5.1.3
The limit superior and limit inferior
Let (xn ) be a bounded sequence. For each n ∈ N, let
yn = sup xm = sup{xn , xn+1 , . . .}.
m≥n
Then {yn } is decreasing, and it is bounded since (xn ) is bounded. By Proposition
5.12, (yn ) converges. The limit is called upper limit of (xn ), and is denoted by
lim sup xn . Similarly, let zn = inf m≥n xm = inf{xn , xn+1 , . . .}. Then (zn ) is increasing, and it is bounded since (xn ) is bounded. The limit of (zn ) is called lower
limit of (xn ), and is denoted by lim inf xn . If (xn ) is not bounded from above, then
its upper limit is equal to ∞ and if (xn ) is not bounded from below, then its lower
limit is equal to −∞. Summarizing
lim sup xn = lim xn = inf sup xk = lim sup xk
n≥m k≥n
n→∞ k≥n
lim inf xn = lim xn = sup inf xk = lim inf xk .
n≥m k≥n
n→∞ k≥n
The basic properties of the upper and the lower limits are listed in the following
proposition:
Proposition 5.16. If (xn ) and (yn ) are sequences of real numbers, then:
(a) lim sup(−xn ) = − lim inf xn and lim inf(−xn ) = − lim sup xn .
(c) lim sup(axn ) = a lim sup xn and lim inf(axn ) = a lim inf xn for any a > 0.
(d) lim sup(xn +yn ) ≤ lim sup xn +lim sup yn and lim inf xn +lim inf yn ≤ lim inf(xn +
yn ).
(e) lim inf xn ≤ lim sup xn , with equality if and only if (xn ) converges. In this
case lim sup xn = lim xn .
Theorem 5.17 (Bolzano-Weierstrass Theorem).
Let (xn ) be a bounded sequence in R. Then (xn ) has convergent subsequence.
Proof. By Proposition 5.14, (xn ) has a monotone subsequence (xnk ). Since (xnk )
is bounded, Proposition 5.12 implies that is converges.
Theorem 5.18 (Cauchy’s criterion). A sequence (xn ) of real numbers converges
if and only if (xn ) is a Cauchy sequence.
Proof. We already know that a convergent sequence is Cauchy (Proposition 5.7).
Hence we have to prove that a Cauchy sequence (xn ) converges to some x ∈ R. By
Proposition 5.7, (xn ) is bounded so that by Theorem 5.17, it contains a convergent
subsequence, say xnk → x. We claim that xn → x. To see this, take ε > 0. Then
there is K such that
|xnk − x| < ε/2 for all k ≥ K.
26
Moreover, since (xn ) is Cauchy, there is N such that
|xn − xm | < ε/2 for all n, m ≥ N .
Take k such that nk ≥ N and let n ≥ N . Then
|xn − x| ≤ |xn − xnk | + |xnk − x| < ε/2 + ε/2 = ε.
Hence (xn ) converges to x, as claimed.
Example 5.19. Define (xn ) by setting x1 = 1, x2 = 2, and
xn =
1
(xn−1 + xn−2 ) .
2
We claim that the sequence converges to x = 5/3. To see this we shall show that
(xn ) is Cauchy and that the subsequence (x2n+1 ) converges to x = 5/3. First note
that 1 ≤ xn ≤ 2 for all n. Next
|xn − xn+1 | =
1
.
2n−1
1
Indeed, for n = 1, |x1 − x2 | = 1 = 210 . Assuming that |xn − xn+1 | = 2n−1
, we show
that |xn+1 − xn+2 | = 21n . We have
1
1
1
|xn+1 − xn+2 | = xn+1 − (xn+1 − xn ) = |xn − xn+1 | = n ,
2
2
2
as claimed. Consequently, if m > n,
|xn − xm | ≤ |xn − xn+1 | + |xn+1 − xn+2 | + . . . + |xm−1 − xm |
1
1
1
≤ n−1 + n + . . . + m−2
2
2
2
1
1
1
1
= n−1 1 + + . . . + m−n−1 < n−2
2
2
2
2
and since 1/2n−2 → 0, it follows that indeed (xn ) is Cauchy. Now to find the limit
of x we first show that x2n−1 < x2n . This holds true for n = 1. Assuming that
x2n−1 < x2n , we show that x2n+1 < x2n+2 . From the assumption if follows that
x2n−1 <
1
(x2n + x2n−1 ) < x2n .
2
So, x2n−1 < x2n+1 and
x2n+2 =
1
1
(x2n+1 + x2n ) > (x2n−1 + x2n ) = x2n+1 ,
2
2
as claimed. From this and |x2n+1 − xn+2 | = 1/22n , it follows that
x2n+2 = x2n+1 +
27
1
.
22n
Now use induction to prove that
x2n+1 = 1 +
1
1
1
+ . . . + 2n−1 .
+
2 23
2
For n = 1 this is clear. Assuming the above equality we show that x2n+3 =
1
. We have
1 + 21 + 213 + . . . + 22n+1
1
1
1
x2n+3 = (x2n+2 + x2n+1 ) =
+ 2x2n+1
2
2 22n
1
1
1
1
= 1 + + 3 + . . . + 2n−1 + 2n+1 ,
2 2
2
2
as claimed. Now, note that
1
1
1
+ 3 + . . . + 2n−1
2 2
2
1
1
1
1
1 + 2 + 4 + . . . + 2n−2
=1+
2
2
2
2
1
1
1
1
1 + + 2 + . . . + n−1
=1+
2
4 4
4
1
5
2
1− n → ,
=1+
3
4
3
x2n+1 = 1 +
where we have used the formula
Pn
k=0
ak =
28
1−an+1
1−a
provided that a 6= 1.