1. Chapter 4
Sequences

2. Section 4.2
Limit Theorems

3. 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) = s t
(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 |
Thus, lim (sn + tn) = s + t.
The proof of (b) is Exercise 4.

4. 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) = s t
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 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
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
Thus, lim sn tn= s t.

5. 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
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
by Exercise 3.2.6(a).
There also exists N2  such that n  N2 implies that
Let N = max {N1, N2}. Then n  N implies that
Hence lim (1/ tn)= 1/ t. 

6. Example 4.2.2* Show that This is Example 4.1.9* from the last section.5
We have sn = 8
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.
And,

7. 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  s
Thus there exists N1  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). 

8. 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 implies L  0. Suppose L < 1.
Then there exists a real number c such that 0  L < c < 1.
Then since (sn + 1/sn) converges to L, there exists N 
such that n  N implies that
Let  = c – L so that  > 0.
Let k = N + 1. Then for all n  k we have n – 1  N, so that
It follows that, for all n  k,
Letting M = sk /ck, we obtain 0 < sn < M cn for all n  k.
Since 0 < c < 1, Exercise 4.1.7(f ) implies that lim cn = 0.
Thus lim sn = 0 by Theorem 

9. 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
Show that
This time we need a lower bound on the numerator.
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.
Thus for n  2 we obtain
To make 3n/2 > M, we want n > 2M /3.
So given any M  , take N > max {2, 2M / 3}.
The formal proof is in the text.

10. We conclude with two theorems for infinite limits
We conclude with two theorems for infinite limits. The proofs are left for the exercises.
Theorem
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
Let (sn) be a sequence of positive numbers. Then
lim sn = +  if and only if lim (1 / sn) = 0.