10
INFINITE SERIES
10.1 Sequences
Preliminary Questions
1.
What is
a
4
for the sequence
a
n
=
n
2
−
n
?
SOLUTION
Substituting
n
=
4 in the expression for
a
n
gives
a
4
=
4
2
−
4
=
12
.
2.
Which of the following sequences converge to zero?
(a)
n
2
n
2
+
1
(b)
2
n
(c)
−
1
2
n
SOLUTION
(a)
This sequence does not converge to zero:
lim
n
→∞
n
2
n
2
+
1
=
lim
x
→∞
x
2
x
2
+
1
=
lim
x
→∞
1
1
+
1
x
2
=
1
1
+
0
=
1
.
(b)
This sequence does not converge to zero: this is a geometric sequence with
r
=
2
>
1; hence, the sequence diverges
to
∞
.
(c)
Recall that if

a
n

converges to 0, then
a
n
must also converge to zero. Here,
−
1
2
n
=
1
2
n
,
which is a geometric sequence with 0
<
r
<
1; hence,
(
1
2
)
n
converges to zero. It therefore follows that
(
−
1
2
)
n
converges
to zero.
3.
Let
a
n
be the
n
th decimal approximation to
√
2. That is,
a
1
=
1,
a
2
=
1
.
4,
a
3
=
1
.
41, etc. What is
lim
n
→∞
a
n
?
SOLUTION
lim
n
→∞
a
n
=
√
2.
4.
Which sequence is defined recursively?
(a)
a
n
=
2
+
n
−
1
(b)
b
n
=
4
+
b
n
−
1
SOLUTION
(a)
a
n
can be computed directly, since it depends on
n
only and not on preceding terms. Therefore
a
n
is defined explicitly
and not recursively.
(b)
b
n
is computed in terms of the preceding term
b
n
−
1
, hence the sequence
{
b
n
}
is defined recursively.
5.
Theorem 5 says that every convergent sequence is bounded. Which of the following statements follow from Theorem
5 and which are false? If false, give a counterexample.
(a)
If
{
a
n
}
is bounded, then it converges.
(b)
If
{
a
n
}
is not bounded, then it diverges.
(c)
If
{
a
n
}
diverges, then it is not bounded.
SOLUTION
(a)
This statement is false. The sequence
a
n
=
cos
π
n
is bounded since
−
1
≤
cos
π
n
≤
1 for all
n
, but it does
not converge: since
a
n
=
cos
n
π
=
(
−
1
)
n
, the terms assume the two values 1 and
−
1 alternately, hence they do not
approach one value.
(b)
By Theorem 5, a converging sequence must be bounded. Therefore, if a sequence is not bounded, it certainly does
not converge.
(c)
The statement is false. The sequence
a
n
=
(
−
1
)
n
is bounded, but it does not approach one limit.
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S E C T I O N
10.1
Sequences
1133
Exercises
1.
Match the sequence with the general term:
a
1
,
a
2
,
a
3
,
a
4
, . . .
General term
(a)
1
2
,
2
3
,
3
4
,
4
5
, . . .
(i) cos
π
n
(b)
−
1
,
1
,
−
1
,
1
, . . .
(ii)
n
!
2
n
(c) 1
,
−
1
,
1
,
−
1
, . . .
(iii)
(
−
1
)
n
+
1
(d)
1
2
,
2
4
,
6
8
,
24
16
. . .
(iv)
n
n
+
1
SOLUTION
(a)
The numerator of each term is the same as the index of the term, and the denominator is one more than the numerator;
hence
a
n
=
n
n
+
1
,
n
=
1
,
2
,
3
,
. . . .
(b)
The terms of this sequence are alternating between
−
1 and 1 so that the positive terms are in the even places. Since
cos
π
n
=
1 for even
n
and cos
π
n
= −
1 for odd
n
, we have
a
n
=
cos
π
n
,
n
=
1
,
2
,
. . . .
(c)
The terms
a
n
are 1 for odd
n
and
−
1 for even
n
. Hence,
a
n
=
(
−
1
)
n
+
1
,
n
=
1
,
2
, . . .
(d)
The numerator of each term is
n
!
, and the denominator is 2
n
; hence,
a
n
=
n
!
2
n
,
n
=
1
,
2
,
3
,
. . . .
2.
Let
a
n
=
1
2
n
−
1
for
n
=
1
,
2
,
3
,
. . . .
Write out the first three terms of the following sequences.
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 Spring '09
 Carnegie
 Calculus, Infinite Series, lim, →∞

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