Math 152 Class Notes
October 22, 2014
10.1 Sequences
A sequence is a ordered list of numbers:
a1 , a2 , a3 , . . . , an , . . .
Notice that for every positive integer n there is a corresponding number an and so a
sequence can be dened as a function whose domain is the set of integers. Thus the
sequence {a1 , a2 , a3 , . . .} is also denoted by
or
{an }
{an }∞
n=1
Example 1. Presentation of sequences.
n n o∞
(a)
n + 1 n=1
n (−1)n n o∞
(b)
n=0
3n
n√
o∞
(c) n − 3
n
an =
n+1
(−1)n n
an =
3n
an =
n=3
√
n − 3, n ≥ 3
n1 2 3
o
n
, , ,...,
,... ;
2 3 4
n+1
n
o
1 2
3
(−1)n n
0, − , , − , . . . ,
,... ;
3 9 27
3n
n
o
√ √
√
0, 1, 2, 3, . . . , n − 3, . . . ;
Denition. A sequence {an} has the limit L and we write
lim an = L
n→∞
or
an → L as n → ∞
if we can make the terms as close to as we like by taking suciently large.
If lim an exists, we say the sequence converges (or is convergent). Otherwise, we
n→∞
say the sequence diverges (or is divergent).
Calculating Limits
• If {an } and {bn } are convergent sequences and c is a constant, then
lim (an − bn ) = lim an − lim bn
lim (an + bn ) = lim an + lim bn
n→∞
n→∞
n→∞
n→∞
lim can = c lim an
n→∞
n→∞
lim c = c
n→∞
n→∞
lim an
an
n→∞
=
if lim bn 6= 0
lim
n→∞
n→∞ bn
lim bn
lim (an bn ) = lim an · lim bn
n→∞
n→∞
n→∞
n→∞
n→∞
• Suppose a sequence {an } is dened by a function f (x), i.e. an = f (n) for integer n.
If lim f (x) = L, then lim an = L.
x→∞
n→∞
• If f (x) is a continuous function and lim an = L, then lim f (an ) = f (L).
n→∞
n→∞
1
1
=
0
because
lim
= 0.
x→∞ xr
n→∞ nr
Example 3. Given r > 0, lim
Example 4. Determine whether the sequence an = (−1)n is convergent or divergent.
Example 5. Find lim
r
n→∞
Example 6. Find lim √
n→∞
4n + 3
.
7n + 6
n
.
10 + n
√
√
Example 7. Find lim ( n + 5 − n).
n→∞
Example 8. Find lim (ln(3n + 1) − ln(4n)).
n→∞
π
n
Example 9. Find lim sin( ).
n→∞
n2
Example 10. Find lim arctan( 2
).
n→∞
n +1
ln n
.
n→∞ n
Example 11. Find lim
Example 12. For what values of r is the sequence rn convergent?
(−2)n + 1
.
n→∞
32n
Example 13. Find lim
•
Squeeze Theorem for sequences
If an ≤ bn ≤ cn for n ≥ n0 and lim an = lim cn = L, then lim bn = L.
n→∞
n→∞
n→∞
In particular, the following useful fact about limits of sequences follows from the
Squeeze Theorem because −|an | ≤ an ≤ |an |.
•
Theorem. If n→∞
lim |an | = 0, then lim an = 0.
n→∞
(−1)n n2
Example 14. Find lim 3
.
n→∞ n + 1
sin n
.
n→∞ n3
Example 15. Find lim
Monotonic and bounded sequences
Denition. A sequence {an} is called increasing if an < an+1 for all n ≥ 1, that is
a1 < a2 < a3 < · · · .
It is called decreasing if an > an+1 for all n ≥ 1 that is a1 > a2 > a3 > · · · .
A sequence is monotonic if it is either increasing or decreasing.
Example 16. an = n is increasing and divergent.
Example 17. an =
3
is decreasing.
n+5
Denition. A sequence {an} is bounded above if there is a number M such that
an ≤ M for all n ≥ 1.
It is called bounded below if there is a number m such that an ≥ m for all n ≥ 1.
If it is bounded above and below, then {an } is a bounded sequence.
Example 18. an =
n
n
is bounded because 0 <
< 1.
n+1
n+1
Example 19. an = (−1)n is bounded because −1 ≤ an ≤ 1. But it is divergent.
Example 20. Consider the sequence dened by the recurrence relation
a1 = 2,
an+1 = 1 −
1
an
for n = 1, 2, 3, . . .
(a) Find the rst 5 terms of the sequence.
(b) Find the limit of the sequence.
We have seen that not every bounded sequence is convergent and not every monotonic
sequence is convergent. But if a sequence is both bounded and monotonic, then we
have the following theorem.
Monotonic Sequence Theorem. Every bounded monotonic sequence is convergent.
The proof of this theorem is based on the Completeness Axiom for the set of real
numbers which is beyond the scope of this course. The following gure provides a
intuitive illustration for the Monotonic Sequence Theorem.
Example 21. The sequence below is bounded and increasing. Find the limit of the
sequence.
a1 = 2,
1
an+1 = (an + 6).
2