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Chapter 4
Sequences
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 1
Section 4.2
Limit Theorems
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 2
To simplify our work with convergent sequences, we prove several useful theorems
in this section. The first theorem shows that algebraic operations are compatible
with taking limits.
Theorem 4.2.1
Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t.
Then (a) lim (sn + tn) = s + t
(b) lim (ksn) = ks and lim (k + sn) = k + s, for any k 
.
(c) lim (sn tn) = st
(d) lim (sn /tn) = s/t, provided that tn  0 for all n and t  0.
Proof: (a) For all n 
we have
|(sn + tn) – (s + t)| = |(sn – s) + (tn – t)|  |sn – s| + |tn – t| by the triangle inequality.
Given any  > 0,
since sn  s, there exists N1 
such that |sn – s| <  /2 for all n  N1.
Likewise, there exists N2 
such that |tn – t| <  /2 for all n  N2.
Now let N = max {N1, N2}. Then for all n  N we have
|(sn + tn) – (s + t)|  |sn – s| + |tn – t| <      .
2 2
Thus, lim (sn + tn) = s + t.
The proof of (b) is Exercise 4.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 3
Theorem 4.2.1
Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t.
Then (c) lim (sn tn) = st
Proof:
This time we use the inequality
| sn tn – st | = |(sn tn – sn t) + (sn t – st)|
 |(sn tn – sn t)| + |(sn t – st)| = | sn |  | tn – t | + |t|  |sn – s|
We know from Theorem 4.1.13 that the convergent sequence (sn) is bounded.
Thus there exists M1 > 0 such that | sn |  M1 for all n. Letting M = max {M1, | t |},
we obtain the inequality sn tn  st  M tn  t  M sn  s .
Given any  > 0, there exist natural numbers N1 and N2 such that
| tn – t | <  /(2M) when n ≥ N1 and | sn – s | <  /(2M) when n ≥ N2.
Now let N = max {N1, N2}. Then n  N implies that
  
  
    .

M

sn tn  st  M tn  t  M sn  s < M 

 2M 
2 2
 2M 


Thus, lim sn tn= st.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 4
Theorem 4.2.1
Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t.
Then (d) lim (sn /tn) = s/t, provided that tn  0 for all n and t  0.
Proof:
Since sn /tn = sn(1/tn), it suffices from part (c) to show that lim (1/tn) = 1/t.
That is, given  > 0, we must make
t  tn
1 1


< 
tn
t
tn t
for all n sufficiently large.
To get a lower bound on how small the denominator can be, we note that, since t  0,
there exists N1 
such that n ≥ N1 implies that | tn – t | < | t | /2. Thus for n  N1 we have
|t |
|t |

by Exercise 3.2.6(a).
2
2
There also exists N2  such that n  N2 implies that tn  t < 12  | t | 2 .
Let N = max {N1, N2}. Then n  N implies that
tn  t  (t  tn )  | t |  t  tn > | t | 
t  tn
1 1
1 t  tn
2 t  tn



<
< .
tn
t
tn t
t
t
| tn |
|t |
Hence lim (1/ tn)= 1/t. 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 5
Example 4.2.2*
5n2  6
5
 . This is Example 4.1.9* from the last section.
Show that lim 2
8n  3n 8
0
5n 2  6
5  6 / n2

We have sn =
2
8  3 /n
8n  3n
5
8
0
Now lim (1/n) = 0, so lim (1/n2) = 02 = 0,
lim (– 6/n2) = (– 6)(0) = 0, and lim [5 – (6 /n2)] = 5.
Likewise, lim (3/n) = 0, so lim [8 – (3/n)] = 8.
5n2  6
5
And, lim 2
 .
8n  3n 8
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 6
Theorem 4.2.4
Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t.
If sn  tn for all n 
, then s  t.
Proof: Suppose that s > t. Then  = (s – t)/2 > 0, and we have 2 = s – t and t +  = s – .

t
Thus there exists N1 

s
such that n ≥ N1 implies that s –  < sn < s + .
Similarly, there exists N2 
such that n ≥ N2 implies that t –  < tn < t + .
Let N = max {N1, N2}. Then for all n ≥ N we have tn < t +  = s –  < sn,
This contradicts the assumption that sn  tn for all n, and we we conclude that s  t. 
Corollary 4.2.5
If (tn) converges to t and tn  0 for all n 
, then t  0.
Proof: Exercise 4(b). 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 7
Theorem 4.2.7
A “ratio test” for convergence
Suppose that (sn) is a sequence of positive terms and that the sequence of ratios
(sn + 1 / sn) converges to L. If L < 1, then lim sn = 0.
Proof: Corollary 4.2.5 implies L  0. Suppose L < 1.
Then there exists a real number c such that 0  L < c < 1.
Let  = c – L so that  > 0. Then since (sn +1/sn) converges to L, there exists N 
such that n  N implies that
sn 1
 L < .
sn
s
Let k = N + 1. Then for all n  k we have n – 1  N, so that n < L    L  (c  L)  c.
sn 1
It follows that, for all n  k, 0 < sn < sn1c < sn2c 2 <  < sk c nk .
Letting M = sk /ck, we obtain 0 < sn < Mcn for all n  k.
Since 0 < c < 1, Exercise 4.1.7(f) implies that lim cn = 0.
Thus lim sn = 0 by Theorem 4.1.8. 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 8
To handle a sequence such as (sn) = (n) where the terms get larger and larger,
we introduce infinite limits.
Definition 4.2.9
A sequence (sn) is said to diverge to + , and we write lim sn = +  if
for every M  there exists a natural number N such that n  N implies that sn > M.
A sequence (sn) is said to diverge to – , and we write lim sn = –  if
for every M  there exists a natural number N such that n  N implies that sn < M.
Example 4.2.11
4n 2  3
  . This time we need a lower bound on the numerator.
Show that lim
n2
We find that 4n2 – 3  4n2 – n2 = 3n2, when n  2.
For an upper bound on the denominator, we have n + 2  n + n = 2n, when n  2.
4n 2  3
3n 2
3n
Thus for n  2 we obtain


. To make 3n/2 > M, we want n > 2M /3.
n2
2n
2
So given any M  , take N > max {2, 2M / 3}.
The formal proof is in the text.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 9
We conclude with two theorems for infinite limits. The proofs are left for the exercises.
Theorem 4.2.12
Suppose that (sn) and (tn) are sequences such that sn  tn for all n 
.
(a) If lim sn = + , then lim tn = + .
(b) If lim tn = – , then lim sn = – .
Theorem 4.2.13
Let (sn) be a sequence of positive numbers. Then
lim sn = +  if and only if lim (1/ sn) = 0.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.2, Slide 10