Difference Equations to Differential Equations — Section 1.2: Sequences
Difference Equations
to
Differential Equations
Section 1.2
Sequences
Recall that a sequence is a list of numbers, such as
1, 2, 3, 4, . . . ,
2, 4, 6, 8, . . . ,
1 2 3
0, , , , . . . ,
2 3 4
1 1 1
1, − , , − , . . . ,
2 4 8
or
1, −1, 1, −1, . . . .
As we noted in Section 1.1, listing the first few terms of a sequence does not uniquely specify
the remaining terms of the sequence. To fully specify a sequence, we need a formula that
describes an arbitrary term in the sequence. For example, the first example above lists the
first four terms of the sequence {an } with
an = n
for n = 1, 2, 3, . . .; the second example lists the first four terms of {bn } with
bn = 2n
for n = 1, 2, 3, . . .; the third example lists the first four terms of {cn } with
cn = 1 −
1
n
for n = 1, 2, 3, . . .; the fourth lists the first four terms of {dn } with
dn =
(−1)n
2n
for n = 0, 1, 2, 3, . . .; and the fifth lists the first four terms of {en } with
en = (−1)n
for n = 0, 1, 2, . . ..
1
Copyright c by Dan Sloughter 2000
2
Sequences
Section 1.2
As indicated in Section 1.1, we are often interested in the value, if one exists, which
a sequence approaches. For example, the sequences {an } and {bn } increase beyond any
possible bound as n increases, and hence they have no limiting value. To visualize what
is happening here, you might plot the points of the sequence on the real line. For both
of these sequences, the plotted points will march off to the right without any upper limit.
Although a limit does not exist in these cases, we usually write
lim an = ∞
n→∞
and
lim bn = ∞
n→∞
to express the fact that the limits do not exist because the terms in the sequence are
growing without any positive bound. On the other hand, if we plot the points of the
sequence {cn }, as in Figure 1.2.1, we see that although they are always increasing (that
is, moving toward the right), nevertheless they never increase beyond 1. Moreover, even
though no term in the sequence is ever equal to 1, we can see that the points become
arbitrarily close to 1. Hence we say that the limit of the sequence is 1 and we write
lim cn = 1.
n→∞
c1
c2
c3 c4 c5
0
1
Figure 1.2.1 The first five values of cn = 1 − n1
Even though they oscillate between positive and negative values, the terms in the sequence
{dn } approach closer and closer to 0 as n increases. Since it is possible to make dn as close
as we like to 0 by taking n suitably large, we may write
lim dn = 0.
n→∞
Finally, for the sequence {en } there are only two points to plot, alternating between 1
and −1. Since the terms of this sequence oscillate between two numbers, and so do not
approach any fixed limiting value, we say that the sequence does not have a limit.
Another approach to visualizing the limiting behavior of a sequence {an } is to plot the
ordered pairs (n, an ) in the plane for some range of values of n. For example, Figure 1.2.2
shows a plot of the points (n, cn ), n = 1, 2, 3, . . . , 50 for the sequence {cn } given above.
Note how the points approach the horizontal line y = 1, indicating, as mentioned above,
that
lim cn = 1.
n→∞
Section 1.2
Sequences
3
1
0.8
0.6
0.4
0.2
10
20
30
40
50
Figure 1.2.2 Plot of (n, 1 − n1 ) for n = 1, 2, 3, . . . , 50
1
0.5
2
4
6
8
10
-0.5
-1
n
Figure 1.2.3 Plot of (n, (−1)
2n ) for n = 0, 1, 2, . . . , 10
Similarly, Figure 1.2.3 shows a plot of the points (n, dn ), n = 0, 1, 2, . . . , 10; here the points
approach the horizontal axis, y = 0, consistent with our claim that
lim dn = 0.
n→∞
Figure 1.2.4 shows a plot of (n, en ), n = 0, 1, 2, . . . , 20. The fact that this sequence does
not have a limit is manifest in seeing the vertical coordinate of the points oscillate between
1 and −1.
As the concept of a limit is fundamental to the understanding of calculus, it is important that we make the notion more concrete than we have so far. That is, we need
to have a formal definition of limit which exactly captures what we have been discussing
intuitively. The idea is that we should say L is the limit of a sequence {an } if for any open
interval I containing L, no matter how small, we can find a point in the sequence beyond
which all values of the sequence lie in I. Graphically, this means that if we start plotting
the points of the sequence, there will come a time when all points from then on will lie
4
Sequences
Section 1.2
1
0.5
10
5
15
20
-0.5
-1
Figure 1.2.4 Plot of (n, (−1)n ) for n = 0, 1, 2, . . . , 20
within the interval I. This idea is formalized in the following definition, where the open
interval I is expressed in the form (L − , L + ) and the idea that all values of the sequence
beyond a certain point are in this interval is expressed by requiring that |an − L| < , that
is, the distance between an and L is less than , for all n > N .
Definition
We say that the limit of the sequence {an } is L, written
lim an = L,
n→∞
if for every > 0 there exists an integer N such that |an − L| < whenever n > N .
Hence to show that the limit of a sequence is a number L, one must show that
for any positive number , it is possible to find an integer N such that the numbers
aN +1 , aN +2 , aN +3 , . . . are all in the interval (L − , L + ). See Figure 1.2.5.
(
aN+1
aN+ 4
L- ε
aN+3
L
aN+2
)
L+ ε
Figure 1.2.5 an in (L − , L + ) for n > N
Example
We will show that
1
= 0.
n→∞ n
To do so, we must show that for any given > 0, we can find an integer N such that
lim
1
−0 <
n
whenever n > N . Now
1
1
−0 = ,
n
n
Section 1.2
Sequences
5
so we need only determine the values of n for which
1
< .
n
Since
1
1
< if and only if n > ,
n
it follows that we may take N to be the largest integer less than or equal to 1 . Then
whenever n > N , we have
1
n> ,
from which it follows that
1
< .
n
This is exactly what we need in order to conclude, by the definition, that
1
= 0.
n→∞ n
lim
The following definition is useful in situations, such as in the previous example, when
we want the largest integer less than or equal to some given value.
Definition
For any real number x, we may define the floor function, denoted bxc, by
bxc = the largest integer less than or equal to x,
(1.2.1)
and the ceiling function, denoted dxe, by
dxe = the smallest integer greater than or equal to x.
(1.2.2)
For example, b5.3c = 5, dπe = 4, b3c = 3, and d3e = 3. With this notation, we could
define N in the previous example by
1
N=
.
Example
We will show that
lim
1
n→∞ 2n
= 0.
This time we must show that for any > 0, we can find an integer N such that
1
−0 <
2n
6
Sequences
whenever n > N . Now
1
1
−0 = n =
n
2
2
Section 1.2
n
1
,
2
so we need to determine the values of n for which
n
1
< .
2
We need to solve this inequality for n. Since n is in the exponent, we may use logarithms
to simplify the inequality. Although we will not provide a careful treatment of logarithms
until Chapter 6, we will assume for the moment some acquaintance with logarithms using
base 10. Now
n
1
<
2
if and only if
n
1
log10
< log10 ().
2
Since
we have
n
1
1
= n log10
,
log10
2
2
n
1
<
2
if and only if
1
n log10
< log10 ().
2
1
Now log10
< 0, so
2
1
n log10
< log10 ()
2
if and only if
n>
Thus if we let
$
N=
log10 ()
.
log10 12
%
log10 ()
,
log10 12
then
1
−0 <
2n
Section 1.2
Sequences
7
whenever n > N . For example, if we take = 0.001, then, to two decimal places,
log10 ()
= 9.97,
log10 12
and so we would have
N = b9.97c = 9.
This N works because, for n > 9,
1
1
1
1
− 0 = n ≤ 10 =
< 0.001.
n
2
2
2
1024
Problem 12 at the end of this section will ask you to generalize the previous example
to show that
lim rn = 0
n→∞
whenever |r| < 1. This is an important fact that we will make use of later.
In this course we will be concerned more with the development of an intuitive understanding of limits and a computational facility with limits than with the formalism of
verifying a specific limit using the above definition. That is not to say that the definition
is unimportant; rather a good grasp of the concept in the definition is important for a full
understanding of much of what we will do in calculus. In fact, mathematicians of the 19th
century arrived at the definition we have stated in their attempts to clarify confusions that
had developed in mathematics since the time of Newton and Leibniz. However, for the
most part these difficulties are beyond the scope of a text such as this one.
We will see that a few basic properties of limits, combined with a few simple limits like
the ones in the previous two examples, will enable us to compute easily a large number
of limits. To begin considering these properties, consider the case where we already know
that
lim an = L
(1.2.3)
n→∞
and we want to compute
lim kan
n→∞
for some constant k 6= 0. Now (1.2.3) tells us that for any > 0, we may find an integer
N such that for n > N ,
.
|an − L| <
|k|
It follows that for n > N ,
|kan − kL| = |k||an − L| < |k|
= .
|k|
But this is what it means to say that
lim kan = kL.
n→∞
(1.2.4)
8
Sequences
Section 1.2
Note that (1.2.4) is obviously true as well when k = 0. Hence we have the following
proposition.
Proposition
If {an } is a sequence for which
lim an = L,
n→∞
then for any constant k we have
lim kan = k lim an = kL.
n→∞
Example
n→∞
(1.2.5)
Since we have already seen that
1
= 0,
n→∞ n
lim
it follows that
350
1
= 350 lim
= (350)(0) = 0.
n→∞ n
n→∞ n
lim
Now suppose we have two sequences {an } and {bn } with
lim an = L
(1.2.6)
lim bn = M.
(1.2.7)
n→∞
and
n→∞
Then (1.2.6) and (1.2.7) tell us that for any > 0, we can find integers N1 and N2 such
that
|an − L| <
2
whenever n > N1 and
|bn − M | <
2
whenever n > N2 . If we let N be the larger of N1 and N2 , then whenever n > N we will
have
|(an + bn ) − (L + M )| = |(an − L) + (bn − M )|
≤ |an − L| + |bn − M |
(1.2.8)
< + = .
2 2
Note that in (1.2.8) we have used the fact, known as the triangle inequality, that for any
real numbers x and y,
|x + y| ≤ |x| + |y|.
(1.2.9)
Thus we have shown
lim (an + bn ) = L + M.
n→∞
Hence we have the following proposition.
(1.2.10)
Section 1.2
Sequences
Proposition
9
If {an } and {bn } are sequences with
lim an = L
n→∞
and
lim bn = M,
n→∞
then
lim (an + bn ) = lim an + lim bn = L + M.
n→∞
Example
n→∞
n→∞
(1.2.11)
We have
lim
n→∞
8
8
1
= lim 4 + lim
= 4 + 8 lim
= 4 + (8)(0) = 4.
4+
n→∞
n→∞
n→∞
n
n
n
Note that in the last example we used the fact that if k is a constant and an = k for
all n, then
lim an = k.
n→∞
This follows immediately from the definition since
|an − k| = 0
for all values of k, and so any integer N will work for any value of .
Again suppose we have two sequences {an } and {bn } with
lim an = L
n→∞
and
lim bn = M.
n→∞
Then we have
lim (an − bn ) = lim an + lim (−bn ) = lim an + (−1) lim bn = L − M.
n→∞
Proposition
n→∞
n→∞
n→∞
n→∞
(1.2.12).
If {an } and {bn } are sequences with
lim an = L
n→∞
and
lim bn = M,
n→∞
then
lim (an − bn ) = lim an − lim bn = L − M.
n→∞
n→∞
n→∞
(1.2.13)
10
Example
Sequences
Section 1.2
We have
n
3
8
1
1
lim
− n = 3 lim
− 8 lim
= (3)(0) − (8)(0) = 0.
n→∞ n
n→∞
n→∞
5
n
5
Note that we have used the result that
lim rn = 0
n→∞
whenever |r| < 0.
We will state three more properties of limits without justifications. Although the
reasoning behind these results is similar to the reasoning of the previous three propositions,
they require a little more care and are best left to a more advanced course.
Proposition
If {an } and {bn } are sequences with
lim an = L
n→∞
and
lim bn = M,
n→∞
then
lim an bn = ( lim an )( lim bn ) = LM.
n→∞
Example
n→∞
n→∞
(1.2.14)
We have
1
lim 2 =
n→∞ n
Proposition
1
lim
n→∞ n
1
lim
n→∞ n
= (0)(0) = 0.
If {an } and {bn } are sequences with
lim an = L
n→∞
and
lim bn = M,
n→∞
then
lim an
an
L
= n→∞
=
,
n→∞ bn
lim bn
M
lim
n→∞
provided L 6= 0 and bn 6= 0 for all n.
Example
We have
3
3
n−3
lim
1
−
1
−
n−3
n = 1.
n
n = n→∞
lim
= lim
= lim
4
4
n→∞
n→∞ 2n + 4
n→∞ 2n + 4
2
2+
lim 2 +
n→∞
n
n
n
(1.2.15)
Section 1.2
Sequences
11
Note that we can apply the previous proposition only when both numerator and denominator have a limit. Hence, in this example, we first divided the numerator and denominator
by n to put the problem in a form to which we could apply the proposition.
Proposition
Suppose {an } is a sequence with
lim an = L.
n→∞
Moreover, suppose p is a rational number, apn is defined for all n, and Lp is defined. Then
lim apn = ( lim an )p = Lp .
n→∞
Example
n→∞
We have
r
lim
n→∞
Example
3
4− =
n
lim
n→∞
lim
n→∞
4−
3 √
= 4 = 2.
n
1
lim
n→∞ n
p
= 0p = 0.
We have
5
23
18 − + 5
n n
Example
r
For any rational number p > 0, we have
1
lim
=
n→∞ np
Example
(1.2.16)
1
1
+ 23 lim 5 = 18 − (5)(0) + (23)(0) = 18.
n→∞ n
n→∞ n
= lim 18 − 5 lim
n→∞
We have
5
6
5
6
lim 4 + 3 − 5
4+ 3 − 5
n→∞
4n5 + 5n2 − 6
4
n
n
n
n
= .
lim
= lim
=
5
4
18
n→∞ 3n + 4n − 18
n→∞
4
18
3
3+ 4 − 5
lim 3 + 4 − 5
n
n
n→∞
n
n
In general, for sequences of the form of the previous example it is useful to divide both
numerator and denominator by the highest power of n which occurs in the denominator.
Example
As another illustration of the idea in the previous example, we have
3
2
1
+ 2− 3
3n2 + 2n − 1
n = 0 = 0.
lim
= lim n n
3
16
n→∞ 2n − 16n
n→∞
2
2− 2
n
12
Sequences
Section 1.2
Definition If lim an exists, we say the sequence {an } converges. If the sequence {an }
n→∞
does not have a limit, we say the sequence diverges.
An important class of divergent sequences are those for which a limit does not exist
either because the terms grow without an upper bound or because they decrease without
any lower bound, as defined in the following definition.
Definition A sequence {an } is said to diverge to infinity if for any real number M there
exists an integer N such that an > M whenever n > N , in which case we write
lim an = ∞.
n→∞
A sequence {an } is said to diverge to negative infinity if for any real number M there exists
an integer N such that an < M whenever n > N , in which case we write
lim an = −∞.
n→∞
Example
Clearly
lim np = ∞
n→∞
for any value of p > 0. For given any M , we need only take
N=
jp
p
k
|M |
to guarantee that an > M whenever n > N .
Example
We have
lim 2n = ∞
n→∞
since, given any M , 2n > M for all n if M ≤ 0 and 2n > M provided
n>
log10 (M )
log10 (2)
if M > 0.
Suppose the sequence {an } diverges and k 6= 0 is a constant. Then the sequence {kan }
must also diverge since if {kan } converged, then the sequence with nth term
1
(kan ) = an
k
would also converge, contradicting our assumption that {an } diverges.
Proposition If the sequence {an } diverges and k 6= 0 is a constant, then the sequence
{kan } also diverges.
Section 1.2
Sequences
13
If the sequence {an } diverges and the sequence {bn } converges, then the sequence
{an + bn } also diverges since, if it converged, then the sequence with nth term
(an + bn ) − bn = an
would also converge, contradicting our assumption that {an } diverges. Similarly, the
sequence {an − bn } diverges.
Proposition If the sequence {an } diverges and the sequence {bn } converges, then the
sequences {an + bn } and {an − bn } both diverge.
Suppose the sequence {an } diverges, the sequence {bn } converges, and
lim bn 6= 0.
n→∞
(1.2.17)
Now (1.2.17) implies that we can find an integer N such that bn 6= 0 for all n > N . So if
the sequence {an bn } converged, then the sequence with, for n > N , nth term,
1
(an bn ) = an
bn
would also converge, contradicting our assumption that {an } diverges. Hence {an bn } must
diverge.
Proposition
If the sequence {an } diverges, the sequence {bn } converges, and
lim bn 6= 0,
n→∞
then the sequence {an bn } diverges
Finally, if the sequence {an } diverges, the sequence {bn } converges, and bn 6= 0 for all
n, then the sequence
an
bn
diverges since, if it converged, the sequence with nth term
bn
an
bn
= an
would also converge, contradicting our assumption that {an } diverges.
Proposition If the sequence {an } diverges, the sequence {bn } converges, and bn 6= 0 for
all n, then the sequence
an
bn
diverges.
14
Example
Sequences
Section 1.2
Consider
1
2
4n + − 2
4n3 + n − 2
n n .
lim
= lim
7
n→∞ 5n2 − 7n
n→∞
5−
n
(1.2.18)
Now
lim 4n = ∞
n→∞
and
lim
n→∞
so
lim
n→∞
2
1
− 2
n n
= 0,
1
2
4n + − 2
n n
= ∞.
Moreover,
lim
n→∞
7
5−
n
= 5.
Thus the numerator in (1.2.18) diverges while the denominator converges. Hence the ratio
diverges. In fact, it should be clear that
3
lim
n→∞
4n + n − 2
= lim
n→∞
5n2 − 7n
1
2
− 2
n n = ∞.
7
5−
n
4n +
Note that in the previous example it was once again useful to divide numerator and
denominator by the highest power of n in the denominator.
Example
We have
15
− 26n3
2
15 − 26n5
n
= −∞.
lim
= lim
13
n→∞ 13 + n2
n→∞
+1
n2
Example The absolute values of the terms of the sequence {(−2)n } grow without bound,
and so the sequence diverges. However, since the terms alternate in sign, the sequence
neither diverges to ∞ nor to −∞.
Monotone sequences
It is sometimes possible to determine that a given sequence converges without explicitly
computing the limit. One important case involves monotone sequences.
Section 1.2
Sequences
15
Definition We say a sequence {an } is monotone increasing if an ≤ an+1 for all n. We
say a sequence {an } is monotone decreasing if an ≤ an+1 for all n. We say a sequence is
monotone if it is either monotone increasing or monotone decreasing.
Now suppose {an } is a monotone increasing sequence. For such a sequence there either
exists a number P such that an ≤ P for all n or there does not exist such an P . In the latter
case, given any real number M , it is then possible to find integer N such that aN > M .
Since the sequence is monotone, it follows that an > M for all n > N , and so the sequence
diverges to infinity. On the other hand, if there does exist a number P such that an ≤ P
for all n, then there in fact exists a number B such that an ≤ B for all n and B ≤ P
for any number P with the property that an ≤ P for all n. The existence of B, known
as the least upper bound of the sequence {an }, is not at all obvious; indeed, the subtle
properties of the real numbers that imply the existence of B were not fully understood
until the middle part of the 19th century. However, given the existence of B, it is easy
to see that given any > 0, there exists a integer N for which aN > B − (if not, then
B − would be an upper bound for the sequence smaller than B). Since the sequence is
monotone increasing and an < B for all n, it follows that
|an − B| <
for all n > N . That is, we have shown that the sequence converges and
lim an = B.
n→∞
Similar results hold for sequences which are monotone decreasing.
Monotone sequence theorem Suppose the sequence {an } is monotone. If the sequence is monotone increasing and there exists a number P such that an ≤ P for all n,
then the sequence converges. If the sequence is monotone increasing and no such number
P exists, then
lim an = ∞.
n→∞
If the sequence is monotone decreasing and there exists a number Q such that an ≥ Q for
all n, then the sequence converges. If the sequence is monotone decreasing and no such
number Q exists, then
lim an = −∞.
n→∞
Example As we shall see in Sections 1.4 and 1.5, we often work with sequences without
having an explicit formula for each term in the sequence. For example, suppose all we
know about the sequence {an } is that a1 = 4 and
an+1 =
1
an
2
16
Sequences
Section 1.2
for n = 1, 2, 3, . . .. That is, the first term in the sequence is 4 and then each successive
term is one-half of its predecessor. Thus
a1 = 4,
a2 = 2,
a3 = 1,
1
a4 = ,
2
and so on. Hence {an } is monotone decreasing. Moreover, every term in the sequence is
positive, so an ≥ 0 for all n. Thus, by the Monotone Sequence Theorem, {an } converges.
Moreover, note that
1
an+1 = an
2
implies that
lim an+1 =
n→∞
1
lim an .
2 n→∞
(1.2.19)
If we let
L = lim an = lim an+1 ,
n→∞
n→∞
then (1.2.19) becomes
L=
1
L.
2
Hence L = 0. That is,
lim an = 0.
n→∞
Problems
1. For each of the following, find a general expression for the nth term of a sequence
which would yield these values as the first four terms.
1 1 1
1 1 1
(a) 1, , , , . . .
(b) 1, , , , . . .
3 9 27
2 3 4
3 5 7
1 1 1 1
(c) 1, , , , . . .
(d) − , , − , , . . .
2 3 4
3 5 7 9
2. For each of the following, decide whether the given sequence converges or diverges. If
the sequence converges, find its limit.
1
, n = 0, 1, 2, . . .
3n
3n − 1
(c) bn =
, n = 1, 2, 3, . . .
2n + 6
3n4 − 6n3 + 1
(e) an =
, n = 1, 2, 3, . . .
5n3 + n2 + 2
(a) an =
(b) an = π n , n = 0, 1, 2, . . .
(d) cn = cos(πn), n = 0, 1, 2, . . .
2n5 − 3n2 + 23
(f) bn = 5
, n = 1, 2, 3, . . .
7n + 13n4 − 12
Section 1.2
(g) cn =
Sequences
45 − 16n2
, n = 1, 2, 3, . . .
13 + 5n + 6n3
(i) an = (−2)2n+1 , n = 1, 2, 3, . . .
r
3n2 + n − 6
(k) an =
, n = 1, 2, 3, . . .
5n2 + 16
17
3n + 1
(h) bn = √
, n = 1, 2, 3, . . .
4n2 + 1
10 − 16n3
(j) an =
, n = 1, 2, 3, . . .
1 + n2
(l) bn =
(−1)n
, n = 0, 1, 2, . . .
5n
3. Explain why
−1 ≤
sin(n)
≤1
n
for n = 1, 2, 3, . . .. What can you conclude about lim
n→∞
sin(n)
?
n
1 n
, n = 1, 2, 3, . . ..
4. Let an = 1 +
n
(a) Compute a1 , a2 , a3 , a4 , and a5 using a calculator.
(b) Compute values of an for n = 1, 2, 3, . . . , 200.
(c) Plot the points (n, an ) for n = 1, 2, 3, . . . , 200, along with the horizontal line y = e.
(d) Does it seem reasonable that lim an = e?
n→∞
(e) What is the smallest value of n for which an > e?
(f) What is the first value of n for which |an − e| < 0.01? Recall that e = 2.71828 to
five decimal places.
1
, n = 1, 2, 3, . . ..
5. Let an = n sin
n
(a) Compute a1 , a2 , a3 , a4 , and a5 using a calculator.
(b) Compute values of an for n = 1, 2, 3, . . . , 200.
(c) Plot the points (n, an ) for n = 1, 2, 3, . . . , 200, along with the horizontal line y = 1.
(d) Does it seem reasonable that lim an = 1?
n→∞
(e) What is the smallest value of n for which an > 0.999?
(f) What is the first value of n for which |an − 1| < 0.0001?
6. Let an = 1.01n and bn = 0.99n for n = 0, 1, 2, . . .. On the same graph, plot the points
(n, an ) and (n, bn ) for n = 0, 1, 2, . . . , 200. How do these two plots compare? Do the
sequences converge?
7. Let an =
10n
for n = 1, 2, 3, . . ..
n!
(a) Plot the points (n, an ) for n = 1, 2, 3, . . . , 100.
(b) From the picture in part (a), can you guess lim an ?
n→∞
(c) What is the maximum value of an for n = 1, 2, 3, . . . , 100?
18
Sequences
Section 1.2
(d) Can you see why
kn
=0
n→∞ n!
lim
for any constant k?
8. Consider the sequence {an } with a1 = 10 and
an+1 =
1
an
3
for n = 1, 2, 3, . . .. Plot the points (n, an ) for n = 1, 2, 3, . . . 50. Do you think this
sequence has a limit? Can you verify this?
9. Consider the sequence {an } with a1 = 2 and
an+1 = 2an
for n = 1, 2, 3, . . .. Plot the points (n, an ) for n = 1, 2, 3, . . . , 50. Can you find the
limit of this sequence using the same method you used in part Problem 8? Does this
sequence have a limit?
10. Consider the sequence {an } with a1 = 0.9 and
an+1 = 2an (1 − an )
for n = 1, 2, 3, . . .. Plot the points (n, an ) for n = 1, 2, 3, . . . , 100. Do you think this
sequence has a limit? If so, can you find it?
11. In each of the following, for an arbitrary > 0, find the smallest integer N for which
|an − L| < whenever n > N . Verify that your value for N works in the particular
case = 0.001.
1
(b) an = 0.98n , L = 0
(a) an = 1 − , L = 1
n
3n3 − 1
1
(c) an = 2 , L = 0
(d) an =
,L=3
n
n3
12. Show that for any −1 < r < 1, lim rn = 0.
n→∞
13. Find sequences {an } and {bn } such that {an } and {bn } both diverge, but {an + bn }
converges.
14. Find sequences {an } and {bn } such that {an } diverges, {bn } converges, and {an bn }
converges.