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10
INFINITE SEQUENCES
AND SERIES
OVERVIEW Everyone knows how to add two numbers together, or even several. But how
do you add infinitely many numbers together? In this chapter we answer this question,
which is part of the theory of infinite sequences and series.
An important application of this theory is a method for representing a known differentiable function f(x) as an infinite sum of powers of x, so it looks like a ''polynomial with
infinitely many terms." Moreover, the method extends our knowledge of how to evaluate,
differentiate, and integrate polynomials, so we can work with even more general functions
than those encountered so far. These new functions are often solutions to important problems in science and engineering.
10.1
HrSlORICAL ESSAY
Sequences and Series
Sequences
Sequences are fundamental to the study of infinite series and many applications of mathematics. We have already seen an example of a sequence when we studied Newton's
Method in Section 4.6. There we produced a sequence of approximations Xn that became
closer and closer to the root of a differentiable function. Now we will explore general sequences of numbers and the conditions under which they converge.
Representing Sequences
A sequence is a list of numbers
in a given order. Each of aJ, a2, a3 and so on represents a number. These are the terms of
the sequence. For example, the sequence
2,4,6,8, 10, 12, ... , 2n, ...
has first term al = 2, second term a2 = 4, and nth term an = 2n. The integer n is called
the index of an, and indicates where an occurs in the list. Order is important. The sequence
2, 4, 6, 8 ... is not the same as the sequence 4, 2, 6, 8 ....
We can think of the sequence
as a function that sends 1 to al , 2 to a2, 3 to a3, and in general sends the positive integer n
to the nth term an. More precisely, an infinite sequence of numbers is a function whose
domain is the set of positive integers.
The function associated with the sequence
2, 4, 6, 8, 10, 12, ... , 2n, ...
sends 1 to al = 2, 2 to a2 = 4, and so on. The general behavior of this sequence is described by the formula an = 2n.
532
10.1 Sequences
533
We can equally well make the domain the integers larger than a given number no, and
we allow sequences of!bis type also. For example, the sequence
12, 14, 16, 18,20,22 ...
is described by the fonnula a" = 10 + 2n. It can also be described by the simpler fonnula
bn = 2n, where the index n starts at 6 and increases. To allow such simpler fonnulas, we
let the first index of the sequence be any integer. In the sequence above, {an} starts with a,
while {b.} starts withb 6 •
Sequences can be described by writing rules that specify their terms, such as
• r
an=vn,
bn = (_I)n+11
n'
dn = (_I)n+1,
n - I
cn=-n-'
or by listing !enns:
{an}
=
{Vi, Vi, v'3, ... , Vn, ... }
{b n }
=
{1,-!,t,-i, .. ·,(-I r1 k, ... }
{cn}
=
1234
n-I
}
{ 0'2'3'4'5'''''-n-''''
{dn }
= {I, -I, I, -I, I, -I, ... ,
(-I).+" ... }.
We also sometimes write
{a,,}
=
{Vn }:,.
Figure 10.1 shows two ways to represent sequences graphically. The fIrst marks the rrrs!
few points from a" a2, a3,"" an,'" on the real axis. The second method shows the graph
of the function defIning the sequence. The function is dermed only on integer inputs, and the
graph consists of some points in the xy-plane located at (I, ad, (2, a2), ... , (n, a,,), ....
-+--~~~I--~I--~
'n
2
3
4
5
FIGURE 10.1 Sequences can be represented as peints on the real line or as
points in the plane where the horizontal axis n is the index number of the
term and the vertical axis all is its value.
Convergence and Divergence
Sometimes the numbers in a sequence approach a single value as the index n increases.
This happens in the sequence
534
Chapter 10: Infinite Sequences and Series
whose terms approach 0 as n gets large, and in the sequence
{o,~, t, ~'~'"'' I
-
k, .. ·}
whose terms approach I. On the other hand, sequences like
{VI, Yz, V3, ... , V;;, ... }
L-E
o
L L+E
•• ••• • (• • -I- )
~------------------- L+E
L
have tenDs that get larger than any number as n increases, and sequences like
------------(n,a,,)- ~ - . ---
L-E
{I, -I, I, -I, I, -I, ... , <-1),+1, ... }
bounce back and forth between I and - I , never converging to a single value. The following definition captures the meaning of having a sequence converge to a limiting value. It
says that if we go far enough out in the sequence, by taking the index n to be larger than
some value N, the difference between a, and the limit of the sequence becomes less than
any preselected number E > O.
DEFINmONS
The sequence {a,} converges to the number L iffor every
positive number E there corresponds an integer N such that for all n,
~~~~----~~~----~n
o
2 3
N
n
la, - LI <
n>N
In the representation ofa
sequence as points in the plane, all ~ L if
y = L is a horizontal asymptote of the
sequence of points (in, a,)}. In this figure,
FIGURE 10.2
E.
Ifno such number L exists, we say that {an} diverges.
If {an} converges to L, we write lim,~oo an = L, or simply a, --+ L, and call L
the limit of the sequence (Figure 10.2).
all the a,/s after aN lie within E of L.
The definition is very similar to the definition of the limit ofa function fix) asx tends
to 00 (limx~oof<x) in Section 2.6). We will exploit this connection to calculate limits of
HISTORICAL BIOGRAPHY
sequences.
Nicole Oresme
(ca. l32()-1382)
EXAMPLE 1
<a)
lim
n--i'OO
Show that
1n =
0
(b) lim k = k
(any constant k)
n~OO
Solution
<a) Let E
>
0 be given. We must show that there exists an integer N such that for all n,
n>N
This implication will hold if (l/n) < E or n > l/E. If N is any integer greater than
I/E, the implication will hold for all n > N. This proves that lim,~oo(I/n) = O.
(b) Let E
>
0 be given. We must show that there exists an integer N such that for alln,
n> N
=)
Ik - kl <
E.
Since k - k = 0, we can use any positive integer for N and the implication will hold.
This proves that lim.~oo k = k for any constant k.
•
EXAMPLE 2
Solution
Show that the sequence {I, -I, I, -1, I, -I, ... , (-1),+1, ... } diverges.
Suppose the sequence converges to some number L. By choosing E = 1/2 in
the definition of the limit, all terms an of the sequence with index n larger than some N
must lie within E = 1/2 of L. Since the number 1 appears repeatedly as every other term
of the sequence, we must have that the number I lies within the distance E = 1/2 of L.
535
10.1 Sequences
M
.. . ..
~~------~--~.
0 123
N
(a)
...
It follows that IL - II < 1/2, or equivalently, 1/2 < L < 3/2. Likewise, the number
- I appears repeatedly in the sequence with arbitrarily high index. So we must also have
that IL - (- I) I < 1/2, or equivalently, - 3/2 < L < - 1/2. But the number L cannot
lie in both of the intervals (1/2, 3/2) and (-3/2, -1/2) because they bave no overlap.
Therefore, no such limit L exists and so the sequence diverges.
Note that the same wgument works for any positive number € smaller than I, not
~~
.
The sequence {vii} also diverges, but for a different reason. Ai; n increases, its terms
become larger than any fixed number. We describe the behavior of this sequence by writing
lim
n~OO
~~----~----~.
o 123 .
N
m
v;. =
00.
In writing infinity as the limit of a sequence, we are not saying that the differences between
the terms an and 00 become small as n increases. Nor are we asserting that there is some number infinity that the sequence approaches. We are merely using a notation that captures the idea
that an eventually gets and stays 1aIger than any fIxed number as n gets 1aIge (see Figure lO.3a).
The terms of a sequence might also decrease to negative infinity, as in Figure 10.3b.
(b)
FIGURE 10.3 <a) The sequence
diverges to 00 because no matter
what number M is chosen, the
terms oftbe sequence after some
index N all lie in the yellow band
above M. (b) The sequence
diverges to - 00 because all terms
after some index N lie below any
chosen number m.
DEFINmON The sequence {an} diverges to infinity iffor every number M
there is an integer N such that for all n 1aIger than N, an > M. If this condition
holds we write
or
Similarly if for every number m there is an integer N such that for all n
bave an < m, then we say {an} diverges to negative infinity and write
>
N we
or
A sequence may diverge without diverging to infinity or negative inrmity, as we saw
in Example 2. The sequences {I, -2, 3, -4,5, -6,7, -8, ... } and {I, 0, 2, 0, 3, O, ... }
are also examples of such divergence.
Calculating Limits of Sequences
Since sequences are functions with domain restricted to the positive integers, it is not surprising that the theorems on limits of functions given in Chapter 2 bave versions for sequences.
THEOREM 1
Let {an} and ibn} be sequences of real numbers, and let A
and B be real numbers. The following rules hold if Iim.~oo an = A and
liml'l_oo b" = B.
1. Sum Rule:
lim.~oo(an
+ bn )
= A
2. Difference Rule:
lim.~oo(an
- bn )
= A - B
3. Constant Multiple Rule:
lim.~oo(k·bn) =
4. Product Rule:
lim.~oo(an· bn )
5. Quotient Rule:
an
Iim.~oob
n
+B
k·B (any number k)
= A· B
A
= Jj
The proof is similar to that of Theorem I of Section 2.2 and is omitted.
536
Chapter 10: Infinite Sequences and Series
EXAMPLE 3
(a)
lim
,,---+00
By combining Theorem I with the limits of Example I, we have:
(_1)n
= -I' lim
11--+ 00
1n
= -1,0 = 0
ConstantMultipJeRuleandExampJela
n - I) = lim (
(b) lim ( -nI -I11 ) = lim I 11--+00
11--+00
11--+00
I = I - 0 = I andExampJela
Difference Rule
lim 11
11--+ 00
Product Rule
- 7 6
(4In 6 )
(d) lim 4 6 n=lim
•
~OO
n
+3
.~OO I
+
-
7
(3In 6 )
0 - 7
=--=-7.
I
Sum and Quotient Rules
+0
•
Be cautious in applying Theorem I. It does not say, for example, that each of the sequences {a.} and {b.} have limits if their sum {an + bn} has a limit. For instance,
{an} = {1,2,3, ... } and {bn} = {-1,-2,-3, ... } both diverge, but their sum
{an + bn} = {O, 0, O, ... } clearly converges to o.
One consequence of Theorem I is that every nonzero multiple of a divergent sequence
{an} diverges. For suppose, to the contrary, that {can} conveIges fur some number c oF o.
Then, by taking k = II c in the Constant Multiple Rule in Theorem I, we see that the sequence
.... .
L
..
f·
-.
b~.~T-; •••~:::f:-i- /I
a;
..
{i·can}
=
{a.}
•• •• -
~~------------~n
o
FIGURE 10.4 TIre terms of
sequence {b.} are sandwiched
betweeo those of {a,,} and {c.},
forcing them to the same
common limit L.
converges. Thus, {ca.} cannot converge uuless {a.} also converges. If {an} does not converge, then {can} does not converge.
The next theorem is the sequence version of the Sandwich Theorem in Section 2.2.
You are asked to prove the theorem in Exercise 109. (See Figure 10.4.)
THEOREM 2-The Sandwich Theorem for Sequences
Let {an}, {bn} , and {cn}
be sequences of real numbers. If a. :5 bn :5 Cnholds for all n beyond some index
N, and if lUnn_oo an
=
lilIln-+oo en
=
L, then limn_ oo b"
=
L also.
An immediate consequence of Theorem 2 is that, if IbnI :5 Cn and cn ..... 0, then
-en ~ hn ~ en. We use this fact in the next example.
b" ---+ 0 because
EXAMPLE 4
(a)
(b)
Since lin ..... 0, we know that
cosn~O
n
.1.n ..... 0
2
k. . . o
(c) (-l) n
because
_1<
because
0<: .1. <:
- 2" -
because
_1n -<: (_l)n1n
n -
COS 11
n
<1.
- n'
111'.
<: 1
- n"
•
The application of Theorems I and 2 is broadened by a theorem stating that applying
a continuous function to a convergent sequence produces a convergent sequence. We state
the theorem, leaving the proof as an exercise (Exercise 110).
THEOREM 3-The Continuous Function Theorem for Sequences
Let {an} be a
sequence of real numbers. If a. ..... L and if / is a function that is continuous at L
and dermed at all an. then /(an) ..... /(L).
537
10.1 Sequences
EXAMPLE 5
Show that Y(n
+
I)/n --+ I.
Solution We know that (n + I)/n --+ 1. Taking f(x) ~
gives V(n + I)/n --+ VI ~ I.
Vx
and L ~ I in Theorem 3
•
2"
EXAMPLE 6
The sequence {I/n} converges to o. By taking an ~ I/n, f(x) ~
and
L = 0 in Theorem 3, we see that 2'/n ~ f(l/n) --+ f(L) ~ 20 ~ I. The sequence {2'/n}
converges to I (Figure 10.5).
•
Using L'Hopital's Rule
o
••
I
I
3
2
•
>"
FIGURE 10.5 As n --+ 00, lin --+ 0 and
21/· --+ 2° (Example 6). The terms of {lin}
are shown on the x-axis; the terms of
{21/.} are shown as the y-values 00 the
graph of f(x) ~ 2".
The next theorem formalizes the connection between Iim,,~oo an and limx~oo f(x). It enahles US to use I'Hllpital's Rule to f'md the limits of some sequences.
THEOREM 4
Suppose that f(x) is a function defined for all x ;;,: no and that
{an} is a sequence of real numbers such that an ~ f(n) for n ;;,: no. Then
~
lim f(x)
lim an
L
x~OO
~
L.
n~OO
Proof Suppose that lim_oo f(x)
her M such that for all x,
~
L. Then for each positive number E there is a numIf(x) - LI <
x>M
E.
Let N be an integer greater than M and greater than or equal to no. Then
n >N
EXAMPLE 7
=>
a.
~
f(n)
and
lao -
LI
~
If(n) - LI <
E.
•
Show that
Solution The function (lnx)/x is defined for all x ;;,: I and agrees with the given
sequence at positive integers. Therefore, by Theorem 4, 1im,,~oo(1n n)/n will equal
limx~oo(lnx)/x if the latter exists. A single application ofl'Hllpital's Rule shows that
lim Inx
x---+OO
~
X
Weconcludethatlim.~oo(lnn)/n ~
lim I/x
1
~
x---+ OO
Q
1
~
O.
•
O.
When we use I'Hllpital's Rule to f'md the limit of a sequence, we often treat n as a
continuous real variable and dllferentiate directly with respect to n. This saves us from
having to rewrite the formula for an as we did in Example 7.
EXAMPLE 8
Does the sequence whose nth term is
an
~ (~)'
n- 1
538
Chapter 10: Infinite Sequences and Series
Solution The limit leads to the indeterminate form 100. We can apply I 'Hllpital 's Rule if
we first change the form to 00 • 0 by taking the natorallogarithm of an:
Ina. = In("" +- 1l)n
Then,
lim
11_00
In an =
lim
11_00
"In("n +- 11)
00·0 form
In(: ~ :)
()form
lim -"-:--;-----"'--
o
1/"
. -2/(,,2 - 1)
= lim -'--'---c--.---'n~OO
-1/,,2
rHilpital's Rule: diffi:rentiate
numerator and denominator.
lim~=2.
n2 - 1
=
11_00
Since
In an --> 2 and f(x) = eX is continuous, Theorem 4 tells us that
an
= elnail ~
e2 .
•
The sequence {an} converges toe 2 •
Commonly Occurring Limits
The next theorem gives some limits that arise frequently.
THEOREM 5
1.
lim
11- 00
The following six sequences converge to the limits listed below:
In"
= 0
n
lim x ' /n = 1
3.
2. lim'%=1
n~OO
(x> 0)
n~OO
S.
lim
11_ 00
(1+'"-)"=e
n
4. limxn=o
n_OO
(lxl
<
1)
n
X
(any x)
6. lim '"--- = 0
lI_
oon !
In Formulas (3) through (6), x remains rlXed as ,,-->
(any x)
00.
Proof The first limit was computed in Example 7. The next two can be proved by taking
logarithms and applying Theorem 4 (Exercises 107 and 108). The remaining proofs are
•
given in Appendix 5.
EXAMPLE 9
(a)
In ~,,2) =
(b)
W
These are examples of the limits in Theorem 5.
2 ~"
--> 2 • 0 =
Formula 1
= ,,2/n = (,,1In),--> (1)2 = I
(c) "{i3;; = 3 1/n(,,1In)
(d)
0
(-t)"--> 0
--> 1·1 =
1
Formula 2
Formula 3 with x
=
3 and Formula 2
Formula 4 with x
=
-!
2
10.1 Sequences
I Factorial Notation
(e)
The notation n! (Un factorial") means
the product I . 2 . 3 ... n of the integers
from I to n. Notice that
(n + I)! = (n + 1). n!. Thus,
4! = 1'2'3'4 = 24 and
5! = 1·2·3·4'5 = 5'4! = 120. We
derme O! to be I. Factorials grow eveo
faster than exponeotiais, as the table
suggests. The values in the table are
rouoded.
It
e"
It!
1
3
5
10
20
148
22,026
4.9 X 108
1
120
3,628,800
2.4 X 10 '8
(n ~ 2)" (1 + --,.2r-~e-2
=
100'
(C) - - + 0
n!
Fonnula 5 with x
=
-2
Fonnula 6 with x
=
100
539
•
Recursive Definitions
So far, we have calculated each a" directly from the value of n. But sequences are often dermed recursively by giving
1. The value(s) of the initial tenn or tenns, and
2. A rule, called a recursion Cormula, for calculating any later tenn from tenns that precede it.
EXAMPLE 10
(a) The statements a, = 1 and a" = a.-1 + 1 for n > 1 define the sequence 1,2,3, ... ,
n, ... of positive integers. With a, = 1, we have a2 = a, + 1 = 2, a3 = a2 + 1 = 3,
and so on.
(b) The statements a, = 1 and a. = n·a.-1 for n> 1 derme the sequence
1,2,6,24, ... , n!,... of factorials. With a, = 1, we have a2 = 2· a, = 2,
a3 = 3'a2 = 6,,,. = 4'a3 = 24, and so on.
(c) The statements a, = 1, a2 = 1, and a.+1 = a" + a.-1 for n > 2 define the sequence 1, 1,2,3,5, ... of Fibonacci numbers. With a, = 1 and a2 = 1, we have
a3 = 1 + 1 = 2, a4 = 2 + 1 = 3, as = 3 + 2 = 5, and so on.
(d) As we can see by applying Newton's method (see Exercise 133), the statements
xo = Iandx.+1 = x. - [(sinx. - x.2 )J(cosx. - lx.)]forn > odefine a sequence
that, when it converges, gives a solution to the equation sin x - x 2 = o.
•
Bounded Monotonic Sequences
Two concepts that play a key role in determining the convergence of a sequence are those
of a bounded sequence and a monotonic sequence.
DEFINmONS
A sequence {a.} is bounded Crom above if there exists a
number M such that a. ,,; M for all n. The number M is an upper bound for
{a.}. If M is an upper bound for {a.} but no number less than M is an upper
bound for {a.}, then M is the least upper bound for {a,,}.
A sequence {a,,} is bounded from below if there exists a number m such that
a. ;" m for all n. The number m is a lower bound for {a,,}. If m is a lower bound
for {a,,} but no number greater than m is a lower bound for {a.}, then m is the
greatest lower bound for {a,,}.
If {a.} is bounded from above and below, the {a.} is bounded. If {a.} is not
bounded, then we say that {a.} is an unbounded sequence.
EXAMPLE 11
(a) The sequence 1,2,3, ... , n, ... has no upper bound since it eventually surpasses
every number M. However, it is bounded below by every real number less than or
equal to I. The number m = 1 is the greatest lower bound of the sequence.
123
n
.
(b) Thesequence Z'3'4"'" n + 1"" lSbounded above by every real number greater
than or equal to 1. The upper bound M = 1 is the least upper bound (Exercise 125).
The sequence is also bounded below by every number less than or equal to
its greatest lower bound
t,
which is
•
540
Chapter 10: Infinite Sequences and Series
I
Convergent .equences are bounded
If a sequence {a,,} converges to the number L, then by detmition there is a number N
such that Ia" - L I < 1 if n > N. That is,
L-l<a,,<L+l
a"
M ~-------------
'.
forn>N.
If M is a number larger than L + 1 and all of the tmitely many numbers aj, a2, ... , aN,
then for every index n we have an :5 M so that {an} is bounded from above. Similarly, ifm
is a number smaller than L - 1 and all of the numbers aj, a2, ... , aN, then m is a lower
bound of the sequence. Therefore, all convergent sequences are bounded.
Although it is true that every convergent sequence is bounded, there are hounded sequences that fail to converge. One example is the bounded sequence {( _1)n+1) discussed
in Example 2. The problem here is that some hounded sequences bounce around in the
band determined by any lower bound m and any upperboundM (Figure 10.6). An important type of sequence that does not behave that way is one for which each term is at least as
large, or at least as small, as its predecessor.
~~-------..----+'
o
m
123
f-"'---"---'.----~---
FIGURE 10.6 Some bounded sequences
bounce around between their bounds and
fail to converge to any limiting value.
DEFINITION
A sequence {an} is nondecreasing if an :5 an+1 for all n. That
is, at ~ a2 ~ a3 ~ .... The sequence is noninc::reasing if an ~ an+ 1 for all n.
The sequence {an} is monotonic ifit is either nondecreasing or nonincreasing.
EXAMPLE 12
(a) The sequence 1,2,3, ... , n, ... is nondecreasing.
123
n
.
d
.
(b) The sequence 2'
3' 4' ... , n + I"" IS non ecreasmg.
111
I
()Th
c
e sequence 1'2' 4' 8"'" 211""
...
18 nonmcreasmg.
(d) The constant sequence 3, 3, 3, ... ,3, ... is both nondecreasing and nonincreasing.
(e) The sequence I, -I, I, -I, I, -I, ... is not monotonic.
_
A nondecreasing sequence that is bounded from above always has a least upper
bound. Likewise, a nonincreasing sequence bounded from below always has a greatest
lower hound. These results are based on the completeness property of the real numbers,
discussed in Appendix 6. We now prove that if L is the least upper bound of a nondecreasing sequence then the sequence converges to L, and that ifL is the greatest lower bound of
a nonincreasing sequence then the sequence converges to L.
y
y~M
M r-----------~--------y~L
... ..
L r-----------~. -~.~.~~
. ...
THEOREM 6-The Monotonic Sequence Theorem
If a sequence {an} is both
bounded and monotonic, then the sequence converges .
'
Proof Suppose {an) is nondecreasing, L is its least upper bound, and we plot the points
(I, all, (2, a2), ... , (n, an), ... in the xy-plane. If M is an upper bound of the sequence, all
~----------------------->x
o
FIGURE 10.7 If the terms ofa
nondecreasing sequence have an upper
bound M, they have a limit L :5 M.
these points will lie on or below the line y = M (Figure 10.7). The line y = L is the lowest such line. None of the points (n, an) lies above y = L, but some do lie above any lower
line y = L - E, if E is a positive number. The sequence converges to L because
(a) a,,:5 L for all values ofn, and
(b) given any E > 0, there exists at least one integer N for which aN > L -
E.
The fact that {a,,} is nondecreasing tells us further that
for all n 2: N.
Thus, all the numbers an beyond the Nth number lie within E of L. Ibis is precisely the
condition for L to be the limit of the sequence {an} .
The proof for nonincreasing sequences hounded from below is similar.
_
10.1 Sequences
541
It is important to realize that Theorem 6 does not say that convergent sequences are
monotonic. The sequence {( -l)'+1Jn} converges and is bounded, but it is not monotonic
since it alternates between positive and negative values as it tends toward zero. What the
theorem does say is that a nondecreasing sequence converges when it is bounded from
above, but it diverges to infinity otherwise.
Exercises 10.1
Find;ng Terms of • Sequence
Each of Exercises 1--6 gives a formula for the nth term an of a sequence {a,}. Find the values of a" a" a" aod /l4.
1. an
I - n
2.
= --,-
n
~ =
3. a, = 2n
I
4. a" = 2
6. a"
=
+ (-I)'
2' - I
----zn-
ten tenns of the sequence.
10. a,
Each positive integer
26. The sequence 0, 1, 1,2,2,3,3,4, ...
Each of Exercises 7-12 gives the fIrst tenn or_ ofa sequence along
with a recursion formula for the remaining terms. Write out the first
7. a,
8. a,
9. a,
Alternating 1's and O's
25. The sequence 1,0, 1,0, 1, ...
I
I
n.
(-1),+1
Cubes of positive integers
divided by pOWOIS of 5
I 8 27 64
125
24. 25' 125' 625' 3125' 15,625''''
Convergence and Divergence
Which of the sequences {a,} in Exercises 27-90 converge, aod which
diverge? Find the limit of each convergent sequence.
n + (-I)'
27. a" = 2 + (0.1)'
28. all =
n
I - 2n
29·a,,=I+2n
= I, a,+1 = a, + (1/2')
= I, a,+1 = aJ(n + I)
= 2, a,+1 = (-1),+1a,,/2
= -2, a,+1 = naJ(n + I)
= a2 = 1, a n+2 = a n+l + an
repeated
30.a,=
2n
+. I1
1- 3 vn
31. a"
=. + ,
32
33
=n'-2n+1
n-I
34. all =
I 2
70 - 4n
36. a,
(-I)' (I -
.a"
I - 5n 4
n
8n
• all
=
n
+3
n2 +5n+6
n'
35. a"
=
Find;ng a Sequence's Formula
In Exercises 13-26, find a formula for the nth term of the sequence.
37. a"
=
13. The sequence 1, -1, 1, -1, 1, .. .
I ~ with alternating signs
39. a"
=
( _1),+1
2n
I
40. a,
=
k)
(2 - i,)(3 + i,)
(-t)"
14. The sequence -1, 1, -1, 1, -1, .. .
1's with alternating signs
41. a"
=
~n ~ I
42. a,
=
I
(0.9)'
alternating signs
43. a"
=
Sin(f + k)
44. all
= n'11' cos (n'11')
Reciprocals of squares
of the positive integers,
with alternating signs
Powers of2 divided by
45•
11.
at
12.
at =
2, a2
=
-1, all +2
15. The sequence I, -4,9, -16,25, ...
16. The sequence I, -
I I
;V9' -
I I
16' 25''''
1 2 2 ' 2 ' 24
17.
+ (-I)'
I
= an+l/all
9' IT' 15' 18' 21' ...
Squares of the positive
integers, with
multiples of3
Integers differing by 2
divided by prndncts of
consecutive integers
(n;;, 1)(1 - k) 38. a,
=
_ sinn
an -
n
3'
48. all =3
n
n
47. a" = 2'
49.
=
a,
=
In(n+l)
Vn
Inn
50. a, = 1n2n
(0.03)'/'
19. The sequence 0, 3, 8, 15,24, ...
Squares of the positive
51. a"
=
8 ' /'
52. a,
20. The sequence -3, - 2, -I, 0, I, ...
integers diminished by I
httegers, beginning with -3
53. a"
=
(I + 'f,)"
54.a,=(I-k)"
21. The sequence I, 5, 9, 13, 17, ...
Every other odd positive
integer
22. The sequence 2,6, 10, 14, 18, ...
Every other even positive
integer
httegers differing by 3
divided by factorials
55. a" = "()"iOn
57. a"
=
(Ii3)'/'
Inn
59. a" = n /'
'
=
56. a, =
%2
58. a,
(n
=
+ 4),/(,+4)
60. a, = Inn - In(n + I)
542
Chapter 10: Infinite Sequences and Series
61.a,=~
63. a,
=
62. a,
(-4)'
n.
=
n!
10 611
= 211 .3"
67. all
=
(krOn')
=1n(1 + k)'
69. all
= (~)'
3n - I
71. all
= (~r
2n + I
'
66.
all
n!
70.a'=(n~S
:s
=(I (10/11)"
74. a, =(9/10)' + (11112)'
76. a, =sinh (In n)
78.
n(l - cask)
80. a, =(3' + 50)1/0
72. a,
all =
82• a,=
75. an
= tanhn
77. an
= 2n
79. a,
=y;, sin ~
81. an
= tan-I n
83. a,
I
=e)'
3 + "IIi"
85. an
= --n-
=(Iny;,n)'
I 3 7 17
a a + 2b
l' 2' S' 12'···· b' a + b , ....
x>O
3",6"
2-no n!
=
1-1n
y;,tan
n2
• 1
_ 1 smli
(Inn)""
• ~
90. an
I
= /,' pdx,
Recursively Deftned Sequences
limit
cz,,+1 =
1
72
+ all
a ll +1
93. al = -4,
a,+1 =
94. al = 0,
a,,+1 =
VS + Za,
V S + Za,
95. al = 5,
96. al = 3,
a,,+1 =
vSa;;
a,,+1
12 -
=
v;;.
I
I
2
+
What is that limit? (Hint: Use part (a) to show that
r,2 - 2 = ±(i/y,)2 and that Yo is not less than n.)
Vi, VI
+
f(x,)
= Xn
-
Xn? -
a. X'o = I,
X n +l = Xn -
b. X'o = I,
X n +l = X n -
c. X'o = 1,
Xn+l = XII -
d. a,
2+1
Vi, VI
b. The fractions Tn = xJYlI approach a limit as n increases.
!'(x )·
n
2
~ =
1
Xn
2: + Xn
tanxn - 1
sec2 XII
1
c. a,
97.2,2+ , 2 + - -,2+
I , ...
2
1
2+-2
2+--
98.
-I,
102... Suppose that f(x) is differentiable for all x in [0, I] and that
f(O) = O. Define sequence {a,} by the rule a, = nf(i/n).
Show that lim,_ooao = 1'(0). Use the result in part (a) to
fmd the limits of the following sequences {a,}.
all + 6
= all + 2
92. at = -1,
or
respectively.
p>1
1 X
In Exercises 91-9S, assume that each sequence converges and fmd its
I
(a + 2b)2 - 2(a + b)' = +1
Do the sequences converge? If so, to what value? In each case,
begin by identifying the functioo f that generates the sequence.
1
2,
2b 2 = -I or + I , then
-
101. Newton'. method The following sequences come from the recursioo formula for Newtoo's method,
vn 2 - 1 - vn 2 +n
=
generally, that if a 2
Xn+l
1/,'1xdx
89. all = 11
91. at
Here the numerators form one sequence, the denominators form
a second sequence, and their ratios form a third sequence. Let XII
and y, he, respectively, the numerator and the denominator of
the nth fractioo r, = xJy,.
•• VerilY that X1 2 - 2Y12 = -I,xi - 2yi = +1 and, more
87.a1J=n-~
= • ,,------;
+,·,+xlI •
Write out enough early terms of the sequence to deduce a general formula for X, that holds for n '" 2.
100. A sequence of rational numhers is described as follows:
73. all
84.a,=~
88. a,
X II +1 =Xl +X2
65. an
all = - - , -
86. a,
Theory and Examples
99. The first tenn of a sequence is Xl = 1. Each succeeding term is
the sum of all those that come hefore it:
n! (Hint: Compare with I/n.)
n
64.
68. a,
= V'32n + 1
VI + Vi,
VI + VI + VI + Vi, ...
=
nln(1
= n(e 1/'
-
I)
+~)
103. Pythagorean triple. A triple of positive integers a, b, and c is
called a Pythagorean triple if a 2 + b 2 = c 2. Let a he an odd
positive integer and let
b=l~2J
and
c=r~21
be, respectively, the integer floor and ceiling for a 2/2.
543
10.1 Sequences
l~J
2' - I
--y-
117. a"
~
119. a"
~ «-I)' + l)(n ~
120. The fl!St tenD of a sequence is XI ~ cos (I). The next terms are
X2 = XI or cos (2), whichever is larger; and X3 = X2 or cos (3),
whichever is larger (farther to the right). In general,
X,+l ~
6
121. a"
a
a. Show that a 2 + b 2 ~ c 2 • (Hint: Let a ~ 2n
band c in tenDs ofn.)
+
I and express
b. By direct calculation, or by appealing to the accompanying
figure, fmd
104. The 11th root of II!
a. Showthat1imn~oo(2mr)1/(2n) ~ I aodhence, using Stirling's
approximatioo (Chapter 8, Additiooal Exercise 32a), that
Test the approximatioo in part (a) for n ~ 40,50, 60, ... , as
far as your calculator will allow.
105... Assuming that lim.~oo(l/n') ~ 0 if c is any positive coo·
stant, show that
b. Prove that limn~oo(l/n') ~ 0 if c is aoy positive coostant
(Hint: IfE ~ 0.001 andc ~ 0.04, how large shouldNbeto
ensure that Ii/n' - 01 < E ifn > N?)
106. The zipper theorem Prove the "zipper theorem" for sequences: If {an} and {b,} both cooverge to L, then the sequence
converges to L.
108. Prove that lim.~ooXl/' ~ I, (x
> 0).
110. Prove Theorero 3.
is bounded.
3n + I
n+T
113. an
=-,
n.
2"3"
(2n + 3)!
112. a" ~ (n
+
2
2an
-
3
126. Uniqueness of least upper bounds Show that if MI and M2
are least upper bouods for the sequence {an}, then Ml ~ M2.
That is, a sequence cannot have two different least upper bounds.
128. Prove that if {an} is a convergent sequence, then to every positive number E there corresponds an integer N such that for all m
andn,
m > N and n > N
~
I .... -
a" 1
<
E.
130. Limits and subsequences If the terms of one sequence appear
in another sequence in their given order, we call the first sequence a subsequence of the second. Prove that if two subsequences of a sequence {a,,} have different limits LI
L2,
then {an} diverges.
*
verge? Give reasons for your answers.
I
116.a,,~n-1i
SI>-
133. Sequence. generated by N_on'. method Newton's method,
applied to a differentiable functioo f(x), begins with a starting
value Xo and coostructs from it a sequence of numbers {x,} that
under favorable circumstances converges to a zero of f. The
recursion formula for the sequence is
Xn+l
I
Which of the sequences in Exercises 115-124 converge, and which di·
115.a,~I-1i
a n +l =
I)!
114. a" ~ 2 - Ii - 2'
I
1,
al =
125. The sequence {II/(II + I)} bas a lea.t upper bound of 1
Show that if M is a number less thao I, then the terms of
{n/(n + I)} eventually exceed M. That is, if M < I there is ao
inreger N such that n/(n + I) > M whenever n > N. Since
n/(n + 1) < I for every n, this proves that I is a least upper
bouodfor {n/(n + I)}.
132. Prove that a sequence {a,} converges to 0 ifaod only if the
quence of absolure values {I an I} cooverges to O.
In Exercises 111-114, detennine if the sequence is monotooic aod if it
~
124.
131. For a sequence {a,,} the tenDs of even index are denored by au
and the terms of odd index by au+ I. Prove that if au ---+ L and
a2k+1 -+ L, then a" -+ L.
107. Prove that lim.~oo 'C",; ~ 1.
111. a,
4"+ 1 + 3"
4'
n+ I
122. an = - n -
129. Uniqueness of limit. Prove that limits of sequences are
uoique. That is, show that if LI aod L, are numbers such that
On -+ LI and all -+ L2, then Ll = L,..
if c is any positive constant.
109. Prove Theorero 2.
~ I +~
123. a" ~
+ I)}.
max {x.. cos (n
127. Is it true that a sequence {a,} of positive numbers must cooverge if it is bouoded from above? Give reasons for your answer.
vi,;! ~ ~ for large values of n.
D b.
I)
=X
f(xn)
ll -
!'(x
)'
lI
.. Show that the recursioo formula for f(x) ~ x 2
caobewrittelIaSxn+l ~ (xn + a/x,)/2.
D b.
-
a, a
> 0,
Starting with Xo ~ I aod a ~ 3, calculate successive tenDs
of the sequence uotil the display begins to repeat. What number is being apprnsimared? Explain.
544
Chapter 10: Infinite Sequences and Series
D 134. Arecursi.. defioi1ionof"./2
Ifyou start with Xl = I anddefme
the subsequent terms of {x,} by the rule x, = X,_I + COSX,_I,
you geoerate a sequeoce that converges rapidly to 1f/2. (a) Try
it. (b) Use the accompaoying figure to explain why the conver-
a. Calculate and theo plot the frrst 25 terms of the sequeoce.
Does the sequeoce appear to be bouoded from above or be-
gence is so rapid.
b. If the sequeoce converges, find an integer N such that
Ia" - L I :5 0.01 for n '" N. How far in the sequeoce do
you have to get for the terms to lie within 0.0001 of L?
low? Does it appear to converge or diverge? If it does cooverge, what is the limit L?
y
135. a, =
COS~_l
o
I
138.
an+l = an
at =
=
1,
(I + 0~5)"
= an + 5"
QII+I
+ (-2)11
140. a"
sinn
sinn
COMPUTER EXPLORATIONS
Use a CAS to perform the followiog steps for the sequeoces in Exercises 135-146.
10.2
136. a" =
137. at = I,
139. an
~¥-~~--J------>x
%
=
n sin
k
Inn
141. a, = --n-
142. a" = "...
143. a, = (0.9999)'
144. a" = (123456)1/,
145. an
=
n 41
8'
I
146. a" = 19"
n.
Infinite Series
An infinite series is the sum of an infmite sequence of numbers
al + a2 + a, + ... + a, + ...
The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infmitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at the result of
summing the first n terms of the sequence and stopping. The sum of the first n terms
Sn =
al
+
a2
+ a3 + ... + all
is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n geta larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in
Section 10.1.
For example, to assign meaning to an expression like
1+
1
1
1
2 + 4 + "8 +
1
16 + ...
we add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.
Partial sum
Value
Suggestive expression
for partial sum
First:
Sl =
1
1
2 - 1
Second:
s2=1+ 21
Third:
1
1
s,=I+-+2
4
3
2
7
4
1
2-2
1
2-4
nth:
1
1
1
s, =1+-+-+···+-2
4
2,-1
2' - 1
2n - 1
2 -n1-1
2 -
10.2
Infinite Series
545
Indeed there is a pattern. The partial sums form a sequence whose nth tenn is
SrI
=
1
2 - 2,,-1.
This sequence of partial sums converges to 2 because 1im,,~00(I/2·-1) = O. We say
"the sum of the infinite series 1
+ !2 + !4 + ... + _1_
+ ...
2.- 1
is 2."
Is the sum of any fmite numberoftenns in this series equal to 2?No. Can we actually add
an infinite number oftenns one by one? No. But we can still define their sum by defming
it to be the limit of the sequence of partial sums as n --> 00, in this case 2 (Figure 10.8). Our
knowledge of sequences and limits enables us to break away from the confines of finite
sums.
1
114
~
o
112
FIGURE 10.8 As the lengths 1, ,/2, '/.,
1/8
2
'I" ... are added one by one, the sum
approaches 2.
HISTORICAL BIOGRAPHY
Blaise Pascal
(1623-1662)
DEFINmONS
Given a sequence of numbers {a.}, an expression of the form
al+a2+ a3+ .. ·+ a.+ .. ·
is an inf"mite series. The number a. is the 11th term of the series. The sequence
{s.} defmed by
is the sequence of partial sums of the series, the number s. being the 11th partial
sum. If the sequence of partial sums converges to a limit L, we say that the series
converges and that its sum is L. In this case, we also write
00
al + az + ... + a. + ... = ~ a. = L .
•- 1
If the sequence of partial sums of the series does not converge, we say that the
series diverges.
When we begin to study a given series al + az + ... + an + .. " we might not know
whether it converges or diverges. In either case, it is convenient to use sigma notation to
write the series as
or
A useful shor1hand
when smnmation
from 1 to 00 is
understood
546
Chapter 10: Infinite Sequences and Series
Geometric Series
Geometric series are series of the form
00
a + aT + ar2 + ... + ar n- t + ...
=
L ar n- 1
11=1
in which a and r are fixed real numbers and a oF
~:o ar". The ratio r can be positive, as in
o. The
I I
(1)"-1 + ...
1-t+~-···+
(-tf' + ....
1+-+-+···+ 2
4
2
series can also be written as
r~I/2,a~1
'
or negative, as in
r~ -1/3,a~
1
If r = I, the nth partial sum of the geometric series is
s"
= a + a(l) + a(I)2 + ... + a(I)"-1 = na,
and the series diverges because lim"~oo s" = ± 00, depending on the sign of a. If r = -I,
the series diverges because the nth partial sums alternate between a and O. If Ir I oF 1, we
can determine the convergence or divergence of the series in the following way:
rSn
=
+ aT + ar2 + ... + arl'l-l
aT + ar2 + ... + arn- 1 + ar1!
Sn - rSn
=
a - ar"
Sn =
a
s.(1 - r) = a(1 - r")
8n =
Multiply s. by r.
Subtract rSn from Sn. Most of
the terms on the right cancel.
Factor.
a(1 - r")
1- r '
(r oF 1).
We can solve for 311 if r =F 1.
If Irl < I, then rn --> 0 as n --> 00 (as in Section 10.1) and s" --> a/(I - r). Iflrl
then Ir"I--> 00 and the series diverges.
Iflrl < 1, the geometric series a
to a/(I - r):
+ ar + ar2 + ... + ar"-l + ...
> I,
converges
00
L ar n - 1 =
11=1
_a_,
1- r
Irl
< I.
If Ir I ;;,: I, the series diverges.
We have determined when a geometric series converges or diverges, and to what
value. Often we can determine that a series converges without knowing the value to which
it converges, as we will see in the next several sections. The formula a/ (I - r) for the sum
of a geometric series applies only when the summation index begins with n = I in the expression ~:"l ar"-l (or with the index n = 0 if we write the series as ~:o ar").
EXAMPLE 1
The geometric series with a
I
I
I
= 1/9 and r = 1/3 is
I (I )"-'
9 + 27 + Sf + ... = ~ 9 3
EXAMPLE 2
00
1/9
I - (1/3)
The series
00
(-I)n5
4"
5
5
5
4
16
64
~--=5--+---+···
n~O
I
6·
•
10.2
is a geometric series with a
a
547
= 5 and r = -1/4. It converges to
5
a
+ (l/4)
1- r
ar - -
Infinite Series
=
•
4.
EXAMPLE 3
You drop a ball from a meters above a flat surface. Each time the ball hits
the surface after falling a distance h, it rebounds a distance rh, where r is positive but less
than 1. Find the total distance the ball travels up and down (Figure 10.9).
--_( -_ _ _
ar2 __ _
Solution
The total distance is
ar 3 - - S
2ar
= a + 2ar + 2ar 2 + 2ar 3 + ... = a + - = all+r
-l-r
-r
This sum is 2ar/ (I - r) .
If a = 6 m and r = 2/3, for instance, the distance is
(a)
1
S
EXAMPLE 4
~
o
o
Solution
o
o
a
+ (2/3)
(5/3)
= 6 1 _ (2/3) = 6 1/3
•
= 30 m.
Express the repeating decimal 5.232323 ... as the ratio of two integers.
From the definition of a decimal number, we get a geometric series
5.232323 ... = 5
+ 12030 + ~ + ~ + .. .
o
(l00)
(l00)
1
( 1
100
23 (
Q
= 5 + 100 1 + 100 +
o
)2 + ... )
a = 1,
r = 1/ 100
1/ (1 - 0.01)
<Il
~
•
<I>
.,.,
-..
.,.,
., .,A
., .,
.,
.'"
23 ( 1 )
23
= 5 + 100 0.99
518
= 5 + 99 = 99
•
Unfortunately, formulas like the one for the sum of a convergent geometric series are rare
and we usually have to settle for an estimate of a series' sum (more about this later). The
next example, however, is another case in which we can find the sum exactly.
00
(b)
FIGURE 10.9 (a) Example 3 shows how
to use a geometric series to calculate the
total vertical distance traveled by a
bouncing ball if the height of each rebound
is reduced by the factor r. (b) A
stroboscopic photo of a bouncing ball.
EXAMPLE 5
Find the sum of the "telescoping" series
L ( ~
n=i n n
1)'
Solution
We look for a pattern in the sequence of partial sums that might lead to a formula for Sk. The key observation is the partial fraction decomposition
1
n(n
+ 1)
----
n
n
+
l'
so
1
L--,---''----:n=i n(n + 1)
k
k
,&i
(1
1)
n-n+l
and
Removing parentheses and canceling adjacent terms of opposite sign collapses the sum to
Sk =
We now see that Sk ~ 1 as k ~
00.
1- k
1
+
1.
The series converges, and its sum is 1:
00
L 1
n=i n(n + 1)
= 1
.
•
548
Chapter 10: Infinite Sequences and Series
The nth-Term Test for a Divergent Series
One reason that a series may fail to converge is that its terms don't become small.
EXAMPLE 6
The series
00
L,,+I=I+l+1+ ... +,,+I+ ...
•-1"
1
2
3
"
diverges because the partial sums eventually outgrow every preassigned number. Each
term is greater than I, so the sum of" terms is greater than ".
•
Notice that lim.~oo a. must equal zero if the series ~::'~1 a. converges. To see why, let
S represent the series' sum and s. = a, + a2 + ... + a. the nth partial sum. When n is
large, both s. and S.-1 are close to S, so their difference, a., is close to zero. More formally,
an
= 3 11
-
~
3 11 -1
S- S
=
O.
Difference Rule for
sequences
This establishes the following theorem.
I Caution
Theorem 7 does not say that l:;1 an
converges if tJn ---+ O. It is possible for a
series to diverge when an ---+ 0 .
00
If ~ an converges, then an ----i> O.
THEOREM 7
11=1
Theorem 7 leads to a test for detecting the kind of divergence that occurred in Example 6.
The nth-Term Test for Divergence
00
L a. diverges if lim
11=1
an fails to exist or is different from zero.
lI--i'OO
EXAMPLE 7
The following are all examples of divergent series.
00
(a)
L ,,2 diverges because ,,2 --> 00.
n=1
(b)
00,,+1.
,,+1
L
-,,diverges because - , , - --> I.
n=1
(c)
L (- 1).+1 diverges because lim.~oo( - 1).+1 does not exist.
00
.-1
00
-n
(d) ~ 2n
+
EXAMPLE 8
.
.
-n
5 diverges because lim.~oo 2n
+5
1
= -
2 oF o.
•
The series
IIIIII
II
I
1+-+-+-+-+-+-+···+-+-+···+-+···
224444
2'2'
2'
~
2 terms
4 terms
2" terms
diverges because the terms can be grouped into infinitely many clusters each of which
adds to I, so the partial sums increase without bound. However, the terms of the series
form a sequence that converges to O. Example 1 of Section 10.3 shows that the harmonic series also behaves in this manner.
_
10.2 Infinite Series
549
Combining Series
Whenever we have two convergent series, we can add them term by term, subtract them
term by term, or multiply them by constants to make new convergent series.
If ~an = A and ~bn = B are convergent series, then
THEOREM 8
~(an + bn) = ~a. + ~bn = A + B
1. Sum Rule:
2. Difference Rule:
3. Constant Multiple Rule:
(any number k).
Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section lO.1. To prove the Sum Rule for series, let
~=~+_+
...
+~
~=b,+~+···+~.
Then the partial sums of ~(an + bn) are
Sn = (a, + b,) + (a2 + b2) + ... + (an + bn)
= (a, + ... + an) + (b, + ... + bn)
=
An
+ Bn·
Sinee An --+ A and Bn --+ B, we have Sn --+ A + B by the Sum Rule for sequences. The
proof of the Difference Rule is similar.
To prove the Constant Multiple Rule for series, observe that the partial sums of ~kan
form the sequence
Sn = ka, + ka2 + ... + kan = kia, + a2 + ... + an) = kAn.
•
which converges to kA by the Constant Multiple Rule for sequences.
All corollaries of Theorem 8, we have the following results. We omit the proofs.
1. Every nonzero constant multiple of a divergent series diverges.
2. If ~an converges and ~bn diverges, then ~(an + bn) and ~(a.
diverge.
- bn) both
Remember that ~(an + bn) can converge when ~a. and ~bn both diverge.
= I + I + I + "'and~bn = (-I) + (-I) + (-1) +"'diverge,
whereas ~(an + bn) = 0 + 0 + 0 + ... converges to O.
Caution
Forexample,~an
EXAMPLE 9
00
(8)
L
n- l
Find the sums of the following series.
3n-' _ 1
6n ,
00
=
L
n- l
00
=
L
11=1
(
1
1 )
2n-' - 6n-'
1
00
L
I
2n-' - 11=1 6n-'
I
1
1 - (1/2)
=2- Q =±
5
5
1 - (1/6)
Difference Rule
Geometric series with
a ~ 1 andr ~ 1/2,1/6
550
Chapter 10: Infinite Sequences and Series
(b)
Cons_ Multiple Rule
=
4C - ~1/2»)
Geometric series with a
I, r
=
=
1/2
•
=8
Adding or Deleting Terms
We can add a finite number of terms to a series or delete a rmite number of terms without altering the series' convergence or divergence, although in the case of convergence
this will usually change the sum. If ~:;"~1 a, converges, then ~:;;'ka. converges for any
k> land
00
Lan
n=1
00
+
= at
a2
+ ... +
Conversely, if ~:;"-ka. converges for any It
>
ak-t
+
Lan"
1I=k
I, then ~:::'1 a, converges. Thus,
and
HISTORICAL BIOGRAPHY
Richard Dedekind
(1831-1916)
Reindexing
As long as we preserve the order of its terms, we can reindex any series without altering its
convergence. To raise the starting value of the index h units, replace the 11 in the fonnula
fora, by 11 - h:
00
00
L an = 11-L1+11: an-h =
11 - 1
at
+
a2
+ a3 + ....
To lower the starting value of the index h units, replace the 11 in the fonnula for a. by 11
00
+ h:
00
:L an = 11=1-11:
:L an+h =
11=1
at
+
a2
+
Q3
+ ....
We saw this reindexing in starting a geometric series with the index 11 = 0 instead of the
index 11 = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions.
EXAMPLE 10
We can write the geometric series
1
00
~ 2,-1
1
=
I
1
+ 2 + 4 + ...
as
00
I
~2n'
00
1
~ 2,-5'
11=5
00
or even
1
n~2n+4'
The partial sums remain the same no matter what indexing we choose.
•
10.2 Infinite Series
551
Exercises 10.2
Finding nth Partial Sums
Using the nth-Tenn Test
Iu Exercises l--{j, fmd a formula for the nth partial sum of each series
and use it to fmd the series' sum if the series converges.
Iu Exercises 27-34, use the nth-Term Test for divergence to show that
1. 2+~+~+l+ ... +~+ ...
3
9
27
3,-1
2 ~ + _9_ + _9_ + ... + _9_ + ...
• 100
100'
100'
100'
3.1_
1 + 1 _ 1 +"'+(_1),-1_1_+ ...
2
4
8
2,-1
4.1- 2 +4 - 8 + ... + (-1),-12,-1 + ...
the series is divergent, or state that the test is inconclusive.
00
n
~
n(n + I)
27. ~ n + 10
28. ,-I (n + 2)(n + 3)
00
00
I
30. ~+29. ~~4
11""0 n
11- 1 n + 3
00
,
00
I
31. ~ cos"
32. ~+-+
11"'0 e
n
,-I
34. ~ cosmT
I
I
I
I
5. 2. 3 + 3· 4 + 4· 5 + ... + (n + I)(n + 2) + ...
5
5
5
5
6. 1'2 + 2'3 + 3'4 + ... + n(n + I) + ...
Series with Geometric Terms
Iu Exercises 7-14, write out the fITst few tenDs of each series to show
how the series starts. Then fmd the sum of the series.
00 (-I)'
7. ~-4'
, -0
00 7
(5 + I)
(I + -----sn
(-I)')
~
m (~y
+
+
(~)' + (~)' + .. .
1& (~2y + (~2)' + (~2)' + (~2)' + (~2)6 + ...
Repeating Decimals
Express each of the numbers in Exercises 19-26 as the ratio of two
integers.
19. 0.23 = 0.23 23 23 ...
20. 0.234 = 0.234 234 234 ...
21. 0.7 = 0.7777 .. .
22. O.d = O.dddd ... , where d is a digit
= 0.06666 .. .
24. 1.414
(I"-~I
I)
n
00
37. ~ (In v;;-+!
,-I
(t) + (t)' + (t)' + (t)' + (t)' + ...
23. 0.06
11- 1
00
40. ~
211
16. 1+ (-3) + (-3)2 + (-3)' + (-3)' + .. .
17.
00
35. ~
-
36.~(3
3)
(n + I)'
'-1~'
In Vn)
311
In Exercises 15--18, determine if the geometric series converges or diverges. If a series converges. fmd its sum.
15. I +
Iu Exercises 35-40, fmd a formula for the nth partial sum of the series
and use it to detennine if the series converges or diverges. If a series
converges, ftnd its sum.
11=1
00
11. ~ 211
00
Telescoping Series
38. ~ (tan (n) - tan (n - I))
9. ,-I
~4'
13.
11=0
1.414414414 ...
=
25. 1.24123
=
26. 3.142857
1.24123 123 123 .. .
= 3.142857 142857 .. .
(Vn+4 - Vn+3)
Find the sum of each series in Exercises 41-48.
41
~.
4
. ::1 (4n - 3)(4n + I)
43
~
40n
• ,~1 (2n - 1)'(2n + I)'
45.
00(1
I)
~
Vn - v;;-+!
47
~.
'.;:1
(
42
~
. ::1 (2n
6
- 1)(2n + I)
44.~2n+1
,~1 n'(n
46.
00(1
~
21/,
+ I)'
I)
- 21/(,+ 1)
I
_
I
)
In(n+2)
In(n+l)
48. ~(tan-l (n) - tao-I (n + I))
,-I
Convergence or Divergence
Which series in Exercises 49--{j8 converge, and which diverge? Give
reasons for your answers. If a series converges, f"md its sum.
00(1)'
49. ~
• r.::
11=0
v2
51.
~ (-1)'+1
11=1
00
53. ~ cosmT
11=0
3,
2
00
SO. ~(V2r
n=O
52. ~ ( -1)'+1n
n=l
00
54. ~ cos:'11'
n=O
5
552
Chapter 10: Infinite Sequences and Series
00
,-I
59.
61.
~~
11=-1
83. Show by example that ~(aJb,) may diverge even though
and ~b, converge and no b, equals O.
56. ~ In 3'
00
57.
I
00
e-'"
,-0
55. ~
00
58.
10"
00
2"_ 1
00
1
84. Find convergent geometric series A ~ ~a, and B ~ ~b, that
illustrate the fact that ~a"b. may converge without being equal
toAB.
I
~"' Ixl>
11=0 X
I
I)'
~3"
00 (
60.~
I-Ii
~ I~O'
00
•
00
21'.
85. Show by example that ~(a,/b,) may converge to something
other than A/Beven when A ~ ~a., B ~ ~b. #' 0, and no b,
equals O.
62. ~"1Ft n!
2"+ 3"
63. ~11=1
4"
00
64.
86. If ~a, converges and a, > 0 for all n, can anything be said about
~(l/a,,)? Give reasons for your answer.
+ 4"
~311+411
87. What happens if you add a fmite number of terms to a divergent
series or delete a ftnite number of tenns :from a divergent series?
Give reasons for your answer.
65.~1n(n :1)
11=-1
Geometric Series with a Variable x
In each of the geometric series in Exercises 69-72, write out the fl!St
few terms of the series to fmd a and r, and fmd the sum of the series.
Then express the inequality Ir I < I in terms of x and fmd the values
of x for which the inequality holds and the series converges.
69. ~(-I)'x'
70. ~(-I)'x'"
.-0
(-I)' (
~-2- 3 +
00
88. If ~a" converges and ~b. diverges, can anything be said about
their term-by-term sum ~(a, + b,)? Give reasons for your
answer.
89. Make up a geometric series ~ar"-l that converges to the number
5if
a. a
=2
b. a ~ 13/2.
90. Find the value of b for which
1 + eb + e2JJ + e3b + ... = 9.
91. For what values of r dues the infinite series
,-0
72.
~a"
I
sinx
)'
converge? Find the sum of the series when it converges.
10 Exercises 73-78, find the values of x for which the given geometric
senes converges. Also, fmd the sum of the series (as a function ofx)
for those values of x.
00
00
73. ~2'x'
74. ~(-!)y'"
,-0
,-0
00
92. Show that the error (L - so) obtained by replacing a convergent
geometric series with one of its partial sums s" is ar"/( 1 - r).
93. The accompanying figure shows the first five of a sequence of
squares. The outermost square has an area of 4 m2 • Each of the
other squares is obtained by joining the midpoints of the sides of
the squares before it Find the sum of the areas of all the squares.
I)"(x - 3)'
00 ( - 76. ~
75. ~(-I)'(x + I)'
11=0
11=0
00
00
.-
2
78. ~(lnx)'
77. ~ sin"x
Theory and Examples
79. The series in Exercise 5 can also be written as
~
I
,~I (n + I)(n + 2)
d ~
an ,~_I (n
Write it as a sum beginning with (a) n
(0) n ~ 5.
~
I
+ 3)(n + 4) .
-2, (b) n
~
94. Helga von Kocb's snowflake CIIl"W Helga von Koch's snowflake is • curve of infmite length that encloses a region of fmite
area. To see why thls is so, suppose the curve is generated by
starting with an equilateral triangle whose sides have length 1.
0,
80. The series in Exercise 6 can also be written as
00
5
~ n(n + I)
00
and
~ (n +
5
!)(n
Write it as a sum beginning with (a) n
(0) n ~ 20.
+ 2) .
~
-I, (b) n
~
3,
81. Make up an infinite series of nonzero terms whose sum. is
L I b . -3
c. O.
82. (Continuation o/Exercise 81.) Can you make an write series of
nonzero terms that converges to any number you want? Explain.
.. Find the length L, of the nth curve C, and show that
limn. . . oo L" = 00.
b. Find the area An of the region enclosed by C, and show that
lim,~ooA. ~ (8/5)A , .
LOO O
C2
c,

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