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Unformatted text preview: 7/5 Summary of Chapter Eight Sequences A sequence {an} is an ordered list of numbers a1, a2, a3,   v . The ﬁrst number
is al, the second is a2, and so on. The list is inﬁnite in length, although some
or all of the terms (numbers) in the sequence may be equal. A sequence {an} converges to a number L if, given any tolerance (positive
number) 6, we can ﬁnd some number a N in the list so that all terms beyond a N
lie within 6 of L. In this case, we write limnnoo an 2 L, or an —> L as n —> 00‘ If a sequence does not converge to any number, we say that the sequence
diverges. This can happen in several ways: the aims may approach 00 or ——00,
or the a’ns may not approach anything at all. Example 1
lim (4 — 2 4,
n—aoo 'n,
hm —2n3+n—6 _ _g
Twice 5n3 + 9 — 5’
{n2} diverges,
and {sin(n)} diverges because the numbers sin(n) do not approach any numerical value as n increases
( although sin(n) is always between —1 and 1 If limnnoo an : L and limnnoo bn = K, then n—roo 77,400 lim (an+bn=L+K, lirn (an—bn)=L—K, lim can 2 cL for any number c,
Tin—’00 lim (anbn) = LK, ’I'L—>OO and lim (32> = 5 if each bn e 0 and K 7e 0.
n—roo bn K We call {an} monotone increasing if each an < an“. A number U is an
upper bound for {an} if each an 3 U. A monotone increasing sequence that
has an upper bound must converge (although this result does not tell us what
the limit is). We call {an} monotone decreasing if each an+1 < an. A number L is a lower
bound for {an} if each an 2 L. A monotone decreasing sequence that has a
lower bound. must converge (again, this does not tell us what the limit is). Example 2 Let a1 = x/3 and, for n = 1,2,3, ~«, let an+1 : i/3an. The ﬁrst
few terms of this sequence are 0.4 3(13 2 H 3V3\/;/§, and so on. It is not obvious whether this sequence converges, or what the limit
might be. However, ask two questions. (I) Is the sequence bounded above? Notice that al s3 1.7321, a2 x 2.2795,
a3 is 2.6151 and a3 % 2.8009. Based on this ﬂimsy evidence, make an initial
guess that each term of the sequence is less than 3. Now see if this is true.
Certainly a1 < 3. Now proceed by induction. If, for any n, an < 3, then a3,“ 2 3a,, < 3(3) 2 9 implies that an+1 < 3 also. By induction, an < 3 for each positive integer n, so
3 is an upper bound for {an}. (2) Is the sequence increasing? Again, from our numerical evidence, it ap
pears to be. To see if this is true, notice that, for an < an+1 to be true, we must
have a,L < m. This is true if a3, < 3am or an < 3, which we know to be true
from (1) Since {an} is a monotone increasing sequence that is bounded above, this
sequence converges. , What does this sequence converge to? Let limnnDO an 2 L. Then limn_,oo an+1 =
L also. Now take the limit as n ~> oo in the deﬁning relationship an+1 = m.
We get lim an+1 = 3( lim an).
71"“)00 '71—’00 L: V3L. Solve this to get L = 3. This sequence converges to 3. If you compute some of
the ags, you can see the numbers increasing toward 3. For example, a9 ’5 2.9936. Then lnﬁmte Series A sequence is an inﬁnitely long list of numbers. An inﬁnite series (or just
seies) is an inﬁnite sum 2:021 ak 2: a1 + a2 + The numbers a1, a2, form a sequence, and the inﬁnite series as the sum of all the numbers in this sequence.
The nth partial sum of the series 22:1 a}, is the sum of its ﬁrst it terms: n
sn22ak=a1+a2+a3+H+an
19:1 3 Each 5” is a ﬁnite sum, hence is a number, so 81, 52, $3, is a sequence,
the sequence of partial sums of the series 21:1 ak. As n is chosen larger,
. . . ()0 . 5,, includes more terms of the 1nﬁn1te sum Zkzl ak. We say that the series
22:1 ak converges to a number L if the sequence of partial sums converge to L as n approaches inﬁnity: DO 712 2 LL], : lim 3“: lim E ak.
711—900 71—900 k=1 k=l If the partial sums do not have a ﬁnite limit, then we say that the series 220:1 ak
diverges. Example 3 (I) The harmonic series = 1 + % + % + i +    diverges. (2) Ifr is a given number, the series Zkzl r1“ = r + r2 + r3 +  ~ is called a
geometric series. This series converges if —1 < r < 1, and diverges if r 2 1 or
r g —1. If —1 < r < 1, this series converges to r/(l — r). For example, i3k—3+9+27I 81+... 3 § 8 —8 64 512'4096 ‘1_§5,
k=1 8 i<—§)k— §+25 125' 625 ‘3 _3 and 00 7 19
Z diverges.
Ic=1 The series 22:0 rk : 1 + r + r2 + r3 + . :  is also called a geometric series,
and converges (to 1/(1 — r)) exactly when ~«1 < r < 1. (3) The series 220:1 E1; is called a 13— series, and it converges when p > 1
and diverges when p S 1. When p : 1 this is the divergent harmonic series. For the 19— series, it is generally unclear what the series actually converges to
for a given p > 1. Tests for Convergence or Divergence of Series The following tests help us determine whether a series converges or diverges. Limit Test  If lirnn_>00 an 75 0, then 2:021 an diverges. This test can never be used to show that a series converges. If ,limn_s00 a,L =
O, the series may converge or it may diverge. For example 2:021(1/n) diverges
and 220:1(1 /n2) converges, even though limnnooﬂ = limnnooﬂ/r?) = 0.
Having limn_,00 an 2 0 does not settle the issue of convergence or divergence. Example 4 °° 5n+2
Z n+7 91:1   5 2 _
diverges, because limnnoo 7:35— ... 5 # O. 00 Z cos(n) n=1
diverges, because lirnn_>00 cos(n) does not exist { and hence is not zero The integral test, comparison test, and ratio test are only for positive series
:3le an, in Which each an > 0. Integral Test — Given 22:1 am suppose we can replace n by at in an to get
a function f that is continuous and decreasing for a: Z 1. Then 20m and/1 f(a:)d:t both converge, or both diverge.
Comparison Test — If 0 < an 3 19m then: 00 00
if E an diverges, so does E (3”,
n=1 71.21 and
00 00
if E bn converges, so does E an.
n=1 71:1 If all we know is that Z:le bn diverges, or 22le an converges, no conclusion
can be drawn about the other series.
Ratio Test — Suppose each an > O and . a 1
hm M” = L.
naoo an Then 00
if 0 S L < 1, then 2 an converges,
71:1
and
00
if 1 < L g 00, then Zen diverges.
11:].
But,
if L = 1, no conclusion can be drawn from this test.
Example 5 Consider 230:2 Notice that lining“, m = 0, so this series may converge  we don”t know yet. Use the integral test.“ / 1 dleim 1 dm.
2 mln(m) ‘ r—ioo 2 acln(:c) i
l For this integral, let u = 1n(x). Then clu : ids: and f 1 dQIZ/ldUIIHIUI =1n1n(as). m1n(a:) it
Then
Iim T 1 dd“ lim T 1 also
race 2 azln(m) — r—+oo 2 m1n(a:)
lim [ln(ln(z))]§ = lim (ln(1n(r)) —1n(1n(2))) = 00. Therefore 2:022 m diverges.
Example 6 Consider 220:1 Since —1 < Sin(n) < 1, then 0 < 1 +
sin(n) < 2 and
1 + sin(n) < __2_
10H 10’“
But 220:1 converges (2 times a convergent geometric series ), so by compar ison, 2:021 converges. O< Example 7 Consider 220:1 Notice that 1 1 1
0<——— <———:—.
1/n3+1 Mn?) 713/2 But 2:021 7131/2 converges (19— series with p = 3/2 > 1). Therefore the given
series converges.  Exam 1e 8 Consider °°_ 1 . Here we can write
p n__1 2n+3 0<—1—< 1
6n 2n+3' forn: 1,2,3,.
This inequality is true because 2n+3 < 6n for positive integers n. But :3le 6i”
diverges {1/6 times the divergent harmonic series ), so the given series diverges. Example 9 Consider 220:1 Here n! = l e 2   n, the product of the n!
integers from 1 through n. We cannot use the integral test here, because as! is not deﬁned if a: is not a positive integer. Take the ratio of successive terms: 10“+1
an+1 (n+1)! _ 10”+1 n! 10 > 0
an ” 12'!” _ 10n (n+ 1)! — n+ 1 as'n —> oo.By the ratio test, this series converges. Example 10 Consider 220:1 Take the ratio of successive terms: 2...+1 an+1 (n+1)4 ” 4 2M1 2 1 4 2
z 2 : )
an 2" n + 1 2” 1 + 31 n4 as n —> 00. Since this limit is greater than 1, this series diverges. Estimating the Sum of a Positive Series Suppose we know that 22:, an converges to some number s, and that each
an is a positive number. We want to estimate the sum of this series, to within
whatever tolerance or error is given. The key is to compare 5 with the partial
sum 5”, since we can make 3,, as close as we want to s by choosing n large enough. Let
R41 = s — 3n =(a1+a2+~+an+an+1+~')—(a1+a2+~~+an)
=0n+1+an+2+”' Assume that we can replace n by at in an to get a continuous function f that
decreases as a: ——> 00. By an area argument like that used to derive the integral
test, we get 00 Rm 3/ But, for a positive series, we also have 0 < s  8,, = Rn.
Then 00
0<s—snS/ 71, If we choose ‘n so that f(ac)d:c g c, then we will have 0 < s —— sn 5 6, so 5,,
will be within 6 of 5. Example 11 The series 22:, converges (pm series with p = 3 Suppose
we want to approximate the sum 3 of this series to within 1/100. Let f(.’E) =
1/333. The strategy is to choose n so that /00 idm < 1
n m3 ‘“ 100'
To do this, compute 00 1 7" 1 ’I"
/ ——3d:r : lim / 23—3da: = lim [——x‘2]
n x T—>OO n T—iOO 2 71
. < 1 1 > 1 1
= hm + = <‘  T—)O<) 2T2 2712 For this to be true, we need or n > 50. The smallest positive integer satisfying this inequality is n = 8.
Thus 5 is approximated, to within 1/100, by 8 1
5822—158—
k=1 2 =1 .1.
+
+
l
+ 343 512 = 1.1952, to four decimal places. Alternating Series
A series is alternating if the terms alternate in sign. Such a series is often
written 00
Z(—1)k+1ak = 01 — a2 +043 — a4 + ' ",
k=l
with each ak positive.
Alternating Series Test — 2:1(—1)k+1ak converges if the positive terms ak
decrease to zero as k —> oo. . oo _ lc+1l 1 ' _
Example 12 Zk=1( 1) k converges, because k decreases to zero as k in
creases If an alternating series 22:1(—1)’“+1ak converges to s, then the difference
between 8 and the nth partial sum of the series is less than an+1z
s — snl < an“. If the alls decrease to zero as n increases, this enables us to approximate the
sum of the series by a partial sum to whatever degree of accuracy we want. Example 13 Approximate the sum of 2:021(~1)”+1—1— to within 1/100. n2
This alternating series converges because l/n2 decreases to zero as n in—
creases. Let s be the sum of the series. Then 1
IS — Snl < an+1 : To get sn within 1/100 of 5, choose n so that
1 1 ____.___ <___.
(n+ 1)2 M 100 This will hold if (n + 1)2 2 100, or n + 1 Z 10, so we can choose n = 9. The
sum of the series is approximated by 9
1
59 2 Sam“? = 0.82796
k:=l I with an error of no more than 1 / 100. Absolute and Conditional Convergence In general, a series 220:1 ck may be neither positive nor alternating. None
of the above tests would apply to such a series. We say that 220:1 ck converges absolutely if 22:1 ckl converges. It is pos—
sible for 22:1 ck to converge, but 2:021 [ckl to diverge. In this case we say that
2:11 ck converges conditionally. i Example 14 22:1(—1)k+1% converges conditionally. This alternating series
converges, bat the series of absolute values of its terms is 22:1 %, the divergent
harmonic series. Absolute Convergence Test — If 22:1 ickf converges, then 22:1 ck converges
also. Thus, absolute convergence implies convergence (but convergence does not
imply absolute convergence). The value of this fact is that we can sometimes show that a series 22:, C},
converges by applying tests for positive series to 22:1 ickl. If 22:1 ckl con—
verges, so does 21:1 ck. However, if 22:1 Ick diverges, then we do not know
about 220:1 ck, which may converge conditionally. Example 15 Consider 22:1 This series is not positive, nor is it al—
ternating, since sin(k) is positive sometimes, and negative sometimes, but the
terms do not alternate in sign as k: increases. Notice that 1 sin(k)
S k3/2‘ 0< k3,, sinUc! sitika leg/2., ’ converges. Therefore 2:021 ICE/2 . oo 1 g Loo
Since 216:1 Egg—.2 converges, then Zkzl
converges also. This series is absolutely convergent, therefore convergent. Summary of Convergence Tests 00 For any series E an, lim an 75 0 implies divergence
TL—)OO
71:1 integral test
For positive series — comparison test
ratio test 00 alternating series :(—1)"+lbn  convergence if the b’ns decrease to 0 71:1
absolute convergence implies convergence Power Series A series of the form 22:0 Cn(.’r — a)” = Co + 01(115 * a) + 02(35 '" C02 + ' " is called a power series, with center a, and coefﬁcients c0, c1, (:2, Given
the center and the coefﬁcients, we ask: for what values of so does this series
converge? One can show that there are three possibilities: (1) The series may converge only for r : a. This happens with the series
220:0 nix”, in which a = O and on = n! If as % 0, this series diverges. (2) The series may converge for all as. This happens with Zoo ix” n=0 n! (3) The series may converge for all a: in some interval (a — r, a + r), centered at a, and diverge for as > a + r and for a: < a — r. In case (1), the series is uninteresting. In case (2), the series converges for
all m, and we say that it has inﬁnite radius of convergence. In case (3), we say
that the series has radius of convergence r. In case (3), we call (a — r, a + r) the
open interval of convergence of the power series. We can often determine which case applies to a particular series by using
the ratio test to check for absolute convergence. Example 16 Consider 2:020 The ratio of the absolute value of succes—
sive terms is n! 1
m lml = lml ‘“’ 0 n + 1
as n ——> 00, for all x. This power series converges absolutely for all x, and
therefore also converges for all m. Example 17 Consider Zoo 1 (a: + 2)”. The center of this series is —2. 11:0 n3“
T ake the ratio of the absolute value of successive terms: £m+21u+l (n+1)3“+" _ n 3" (:13 + 2)“+1 — n + 1 3n+1 (a: + 2)”
n 1 1 =——m+2«+§m+2 asn—>oo. From the ratio test, the series converges absolutely if this limit is less than 1.
Therefore, the series converges if élm+2<1, or
m+2 <3. This is the interval ——5 < :17 < 1, of radius 3, centered at —2. The series
converges (absolutely) for —5 < at < 1, and diverges ifs: < —5 or ifs: > 1. This
series has radius of convergence 3. Differentiation and Integration of Power Series Suppose 220:0 cn(x — a)" converges either on the entire real line (that is,
for all m), or for a: in some open interval (a — r, a + r). For such as, the power
series deﬁnes a function of as, and we can write f = 2:10 cn(:c —~ a)”. This
function can be integrated by integrating the power series term by term, as long
as we remain within an interval over which the series converges. Thus, as an
indeﬁnite integral, f(x)dx=ch (x~a)”do:=ch 1 (m—a)”+1+C’
nzo n+1 n=0 1 2 1
_ c :c a ‘ —c a: a +
0( ) I 2 1( ) 3
Further, within an interval over which the series converges, we can differen tiate a power series term by term: C2(:B (1)3 I ~~+C'. mac) '= 2 d1; (one — or)
n=0 : ZmnC’E — a)n—1 = C1 + 262(17  a) + 3C3(m — a)2 + .
11:1 Notice that this series for f’ begins with n = 1, because the derivative of the constant term co in the power series for f is zero.
The radius of convergence of the series for f’ is the same as that for f Example 18 From the geometric series, we know that 1 —.’E 00
anzl+m+$2+m3+x4+~. n=0 M) = 1
for —l < m < 1. Then /f(w)da:=/ 1 dmz—lnil—m]+0 1 —:c
_. _ +1 M _ 2 _ 3
—Z/x”dm—Zn+1mn —x+2m +39: +
71:0 71:0
And
1 00 d 00
I _ _ _ —1 a 2 3
f(30) m —Zﬁx”—ann —1+2ac+3x +490 +~7
n—O n=1
for ~1 < m < 1.
r Example 19 Put r = —a:2 into the geometric series 1—; = 2:0:0 r” to get
1 OO 00
1M2 =Z<—wr =Z<1>“w2”=1—w2+m4~w6+~7
n=0 n=0
for —1 < m < 1. With = 1/(1 + :52), we have
29: 0°
/ = _ 2 __1 'n, 2 2n—1
f (w) —————(1 + W n; > < mm for —1<:I:<1. 10 i
g
i
i
i
g i
g Taylor and Maclaurin Series Expansions of Func—
tions Suppose f 2 22:0 cn(m — a)”. How are the coefﬁcients on related to 1%)?
First notice that f(a) 2 c0, because (cc—a)” = O at x : a for n = 1, 2, 3,  ~ .
Now differentiate the series term by term: f'(93) = imam — a)””1 2 c1 + 2c2(x — a)2 + 303(33 , a)3 + I H‘ 7121 If we put CE 2 a, all terms involving x — a on the right vanish, and we are left with
f’(a) : 01.
Differentiate again:
W) = 2m — 1>cn<m — a)“
n22 = 262 +6C3($ — a) + 1204(x —— 502 + . ...
Again putting a: = a, we get f”(a) : 2C2, so 1
62 = if"(a).
Differentiate a third time:
f”’(x) : :nm — 1)(n — 2)cn(ac w (an—3
n=3 = 6C3 + 2404(95 — a) + 60c5(x —— 502 + . ...
Then f”’(a) = 603, so
._ 1 //I _ 1 III
C3 — 6f (6‘) — if (CL) Continuing in this way, we find that in general
1
_— __ (n)
an — n! f (a), where f (n) (a) is the nth derivative of f (x), evaluated at a: = a, and f(0)(a) =
f (a) as a notational convenience. This means that the power series, which we have called f (ac), can be written
in terms of and its derivatives at a as me) = Z imam — a)“. n=0 The power series on the right is the Taylor series of f about a. When a = 0,
this Taylor series is often called a Maclaurin series. 11 i
i Example 20 Let f = e“. We will write the Maclaurin series for f (so
a = 0).The derivatives of em are all e“, so f(”)(x) : em and f(")(0) = e0 = 1. The Taylor coefﬁcients of em about 0 are 1 1 __ (n) __._ Cn‘nlf (0)—n!‘ The Maclaurin series for f(a:) is
00
1 n_ 1 2 1 3 1 4
20—7551: _1+x+—2—!$ +333: +133 +l. n: It is easy to check that this series converges for all real numbers as. This raises the following question. Suppose f and a are given. If f
can be differentiated as many times as we like at a, then we can write the Taylor
series of f about a: Z %f‘"’(a)(w ~ a)”.
n=0 . But how do we know that this series converges to f in some open interval
about a? The answer is provided by the following. Suppose we are dealing with
a: in some interval a  d 5 cc 3 a + d, and we can ﬁnd a number M that bounds
all derivatives f(”) for :0 in this interval: .f(”)(x)lSMfora~dS$Sa+dandn=1,2;3,”‘ Then the Tavlor series of f converges to f at least for a — d < :c < a + d: f(:1:) 22%f(n)(a)(a:—a)n for a—d<x<a+d. n=0 ' Example 21 Let f(ac) 2 cos(:13) and a : 0. We will expand f(ac) in a Taylor
series about 0.
First, the nth derivative of cos(w) must be either cos(9:), sin(a:), — cos(a:), or
— COS(.’13), depending on n. In any case, lf‘”)(rc)l s 1 on any interval. Therefore the Maclaurin series for cos(x) about 0 converges to
cos(m) for all :10. All we have to do is write this series. Compute N3?) = — 5111(90), f”($) = — 005013),
f”’(m) : sin(m) and f(4)($) : cos(:c). 12 g
i
a
a Then
f’(O) = 0, f”(0) = —1, f”’(O) = 0 and Ma» = 1. Every derivative of cos(ac) of odd order is zero at at = O, and every derivative of
even order is either 1 or —1 at :c : 0. The pattern is: f<2“+1><0) = o, f<2"><0> = (~1)”~ The power series is 00 1 n n 00 (—1)” n where in the series on the right we used the fact that all the odd order derivatives
in the series on the left are zero, and kept only the even order terms. Therefore, for all ac,
1 4 1 6 _oo(~—1)n2n_ 1 2
«so»: Wm ”17;; WC 6735 +.... 71:0 Example 2‘2 Suppose we want the Maclaurin series for cos(3m). We do not have to start from the beginning. Just substitute 3st for a: in the series just
derived:
°° (—1)” 2n 2n
(2n)!3 IE . n=0 Example 23 Suppose we want the Maclaurin series for sin(x). Again, we do
not have to start from the beginning. We have the Maclaurin series for cos(:c), so just diﬁ‘erentiate: ~ _ d __d°°(‘1)" 2n
5111(93)  dm (303(93) — dm 7;) (2n)! 3:
_. _ 00 (—1)” 211—1
_ “2:; (2n)! (2n):r .
Now 2n/(2n)! = 1/(2n —— 1)!, so
 00 "llnH 2711 1 3 1 5 1 7
sm($)::(—2;L—_—1—)—!m :m‘im 71:11 This is correct, but this series is more traditionally written in the equivalent
form ' for all m. 13 Some Applications of Power Series Expansions Often power series expansions are used to approximate such quantities as
integrals. Example 24 Appromimate fol/2 cos(ac3)dsc to within 1/ 100000. We cannot evaluate this integral by the Fundamental Theorem. To approwi
mate it, ﬁrst‘ewpand 003(333) in a power series about 0. To do this, replace a by
m3 in the Maclaurin series for cos(.w) to get 00 “1 n 2n 00 _1 n n
cos(m3) = Z ((273)! ($3) :6 ((271))! x6 . 71:0 11:0 Then 1/2 05 3 * DO (—1)” 1/2906” a:
/0 c (whim—E (2n)!/0 d ~—— i til? [6.1+ 0 Now 1
#1180980 — 0000000846 < 100000. Therefore, with an error of less than 1/100000, the sum of this alternating series is % — £76, or 447/896 (about 0.49888 ). Example 25 Approzrimate [01/ 4 arctan(2x)dcc, with an error of no more than
0.001. One strategy is to expand arctan(293) in a Maclaurin series, then integrate
term by term. We do not need to derive this series from scratch. Begin with 00 1 TL
1 + :32 Z 2(4)an > n=0 which we derived previously (from a geometric series). Recognizing that this is
the derivative of arctan(x), write 1
f 1 + m2 dac — arctan(:1:) + C' : T;(—1)”/w2ndm 14 : i $2n+1.
2 +1 n=0 To evaluate the constant C, put :6 = O to get arctan(0)+C : 0. But arctan(0) :
0, so 0 = 0, and we have °° (4)” 1 1 1
arctan(:n) : 2: 2n + 1x2n+1 : x — 33:3 + 5335‘— ?x7 + . .. 71:0 Replacing 90 by 2m, we have arctan(2x) «_ Z 22n+1$2n+1 n=0 2n + 1
Now 1
1/4 00 (_1)n 1/4
/ arctan(2m)dm = 22n+1/ mgnﬂda:
0 2n + 1 0
n=0
_ i0: (_1)n22n+1
‘ n20 (2n + 1)(2n + 2)42n+2
— 2 23 + 25 27 + ‘ H
(2X42) (3)(4)(44) (5)(6)(46) (7)(8)(48) ’
Now
23
(3)(4)44 = 0.0020  a > 0.001,
but
25 Therefore, the sum of the first two terms of this alternating series approximate
the sum of this alternating series with an error less than 0.001, so the approxi
mation we want is 2 ————————23 — 0 059896
(2X42) ’ <3><4><44> ‘ ' ’ 1/4
/ arctan(2m)da: z
0 to ﬁve decimal places. 15 ééiis._._32aixiiﬁsaﬁﬁgéy, gbasicééxpéc£2?a? ...
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This note was uploaded on 09/26/2008 for the course MA 126 taught by Professor O'neil during the Summer '06 term at University of Alabama at Birmingham.
 Summer '06
 O'NEIL
 Calculus

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