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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 260 
"" 0 "" {TEXT -1 49 "Lesson 19:  Convergence Tests for Infinite Series
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "We have seen what is meant by
 saying that an infnite series " }{XPPEDIT 18 0 "Sum(a_n,n = 1 .. infi
nity);" "6#-%$SumG6$%$a_nG/%\"nG;\"\"\"%)infinityG" }{TEXT -1 219 " co
nverges, with sum s.  (You should review that definition now if you do
 not remember it.)  Unfortunately, there are very few series to which \+
the definition can be applied directly; the most important is certainl
y the " }{TEXT 267 16 "Geometric Series" }{TEXT -1 1 " " }{XPPEDIT 18 
0 "Sum(a*r^(n-1),n = 0 .. infinity);" "6#-%$SumG6$*&%\"aG\"\"\")%\"rG,
&%\"nGF(F(!\"\"F(/F,;\"\"!%)infinityG" }{TEXT -1 45 ".  In the same se
ction, it was shown that if " }{XPPEDIT 18 0 "Sum(a_n,n = 1 .. infinit
y);" "6#-%$SumG6$%$a_nG/%\"nG;\"\"\"%)infinityG" }{TEXT -1 17 " conver
ges, then " }{XPPEDIT 18 0 "proc (a_n) options operator, arrow; 0 end;
" "6#R6#%$a_nG7\"6$%)operatorG%&arrowG6\"\"\"!F*F*F*" }{TEXT -1 3 ".  \+
" }{TEXT 268 41 "The converse of this statement is false: " }{TEXT -1 
2 "if" }{TEXT 271 1 " " }{XPPEDIT 18 0 "proc (a_n) options operator, a
rrow; 0 end;" "6#R6#%$a_nG7\"6$%)operatorG%&arrowG6\"\"\"!F*F*F*" }
{TEXT -1 10 ", it does " }{TEXT 270 3 "not" }{TEXT -1 12 " follow that
" }{TEXT 269 1 " " }{XPPEDIT 18 0 "Sum(a_n,n = 1 .. infinity);" "6#-%$
SumG6$%$a_nG/%\"nG;\"\"\"%)infinityG" }{TEXT -1 93 " converges!  (What
 example have you seen that shows this?)  The idea to bear in mind is \+
that " }{XPPEDIT 18 0 "Sum(a_n,n = 1 .. infinity);" "6#-%$SumG6$%$a_nG
/%\"nG;\"\"\"%)infinityG" }{TEXT -1 18 " will converge if " }{XPPEDIT 
18 0 "a_n;" "6#%$a_nG" }{TEXT -1 165 " goes to 0 \"quickly enough\".  \+
It is not possible to give an exact meaning to \"quickly enough\" in t
he last sentence, but you have been or will be shown in class that " }
}}{EXCHG {PARA 258 "" 0 "" {XPPEDIT 18 0 "Sum(1/n,n = 1 .. infinity);
" "6#-%$SumG6$*&\"\"\"F'%\"nG!\"\"/F(;F'%)infinityG" }{TEXT -1 14 " di
verges, but" }}}{EXCHG {PARA 259 "" 0 "" {XPPEDIT 18 0 "Sum(1/(n^(1+ep
silon)),n = 1 .. infinity);" "6#-%$SumG6$*&\"\"\"F')%\"nG,&F'F'%(epsil
onGF'!\"\"/F);F'%)infinityG" }{TEXT -1 28 " converges for any positive
 " }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }}}{PARA 0 "" 0 "" {TEXT 
-1 5 "(the " }{XPPEDIT 18 0 "p;" "6#%\"pG" }{TEXT -1 1 "-" }{TEXT 272 
6 "series" }{TEXT -1 6 "), so " }{XPPEDIT 18 0 "a_n = 1/n;" "6#/%$a_nG
*&\"\"\"F&%\"nG!\"\"" }{TEXT -1 93 " just fails to be fast enough.  (Y
ou may recall a similar situation with improper integrals.)" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We now study so
me basic tests we can use to determine whether a series converges or d
iverges." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 5 "" 0 "" 
{TEXT -1 83 "_________________________________________________________
__________________________" }}{PARA 4 "" 0 "" {TEXT -1 33 "A. The Inte
gral Test & \020P- Series" }}{PARA 0 "" 0 "" {TEXT -1 83 "____________
______________________________________________________________________
_" }}{PARA 0 "" 0 "" {TEXT -1 500 "There are many occasions where a se
ries is closely related to an integral. In these cases,the integral wi
ll converge or diverge if and only if the corresponding sum converges \+
or diverges. This result is calle the Integral Test. Remember a sequen
ce is  function defined on the domain of natural numbers. If it makes \+
sense to define the same function for all positive real numbers, and t
hat functioin is integrable ( able to be intergrated) , then the integ
ral test will apply. Lets look at an example." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Here is a function, the s
equence it defines, and a plot of the sequence and function." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "f
 := x -> 1/x^(3/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG6
\"6$%)operatorG%&arrowGF(*&\"\"\"F-*$)9$#\"\"$\"\"#F-!\"\"F(F(F(" }}}
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C>F($\"++++]7F($\"+7>FW*)!#6$\"+w\"QT!oF/$\"+tC\\*R&F/$\"+$Q<%>WF/$\"+
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{MPLTEXT 1 0 103 "plot( [f(x), f( floor(x)), f(floor(x+1))], x = 1..12
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l7$Fc\\l$\"3CeR<$HiM8#Fer-Fh\\l6&Fj\\lF\\]lF[]lF[]lFcbm-%+AXESLABELSG6
$Q\"x6\"Q!6\"-%%VIEWG6$;F(Fc\\l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 569 "Note tha
t the blue curve is between the two step functions colored yellow and \+
red. If curve encloses infinite area then so does the larger step func
tion because at all times it is greater than or equal to the curve. On
 the other hand, if the curve encloses a finite area, then so does the
 smaller step function. Simple enough. However, the upper step functio
n is the lower one shifted one unit to the right. In other words, the \+
upper one is the same as the lower one plus one additional term of a1 \+
= 1. Consequently, the series converge or diverge as the integral does
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 51 "Sum( f(k), k = 1..infinity):   % = evalf(value(%));" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*&\"\"\"F(*$)%\"kG#\"\"$\"
\"#F(!\"\"/F+;F(%)infinityG$\"+\\`P7E!\"*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 44 "Int( f(x), x = 1..infinity):   % = value(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$)%\"xG#\"\"$\"\"
#F(!\"\"/F+;F(%)infinityGF." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
51 "Sum( f(k+1), k = 1..infinity): % = evalf(value(%));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#/-%$SumG6$*&\"\"\"F(*$),&%\"kGF(F(F(#\"\"$\"\"#F
(!\"\"/F,;F(%)infinityG$\"+\\`P7;!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 340 "In this case, we can actually comput
e the integral and the sums. Note that all three converge, however the
y are not equal. The value of the integral is between the value of the
 sums from k = 1.. infinity and k = 2..infinity - just as we saw on th
e graph above. Thus the integral tells us about convergence & divergen
ce but not exact values." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 119 "One important res
ult that follows directly from the integral test, is that series of th
e form  1/k^p converge only for " }}{PARA 0 "" 0 "" {TEXT -1 30 "p > 1
, and diverges otherwise." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "assume( q > 0);   p := q + 1;   Int
( 1/x^p, x = 1..infinity ):   % = value(%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"pG,&%#q|irG\"\"\"F'F'" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%$IntG6$*&\"\"\"F()%\"xG,&%#q|irGF(F(F(!\"\"/F*;F(%)infinityG*
&F(F(F,F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "assume( q <= 0
);   p := q + 1;   Int( 1/x^p, x = 1..infinity ):   % = value(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG,&%#q|irG\"\"\"F'F'" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F()%\"xG,&%#q|irGF(F(F(!\"\"/
F*;F(%)infinityGF0" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 62 "Note that integral is finite for p > 1, and finite for \+
p =< 1." }}{PARA 0 "" 0 "" {TEXT -1 83 "______________________________
_____________________________________________________" }}{PARA 4 "" 0 
"" {TEXT -1 28 "B. The Limit Comparison Test" }}{PARA 0 "" 0 "" {TEXT 
-1 83 "_______________________________________________________________
____________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 311 "Another important test is the Limit Comparison Test. A
 simplified version of this test asserts that if the ratio of the term
s of two different sequences have a finite, non-zero limit, then both \+
converge or both diverge. Lets examine why test works from a geometric
 point of view. Consider these three sequences." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 " bk is known to converge
 while  ck is known to diverge. For now we are not sure if  ak converg
es or diverges." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 44 "a := n -> (3 + sqrt(n))/(n^2 - sqrt(n) + 5);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aGR6#%\"nG6\"6$%)operatorG%&arro
wGF(*&,&\"\"$\"\"\"-%%sqrtG6#9$F/F/,(*$)F3\"\"#F/F/F0!\"\"\"\"&F/F8F(F
(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "b := n -> n^(-3/2);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bGR6#%\"nG6\"6$%)operatorG%&ar
rowGF(*&\"\"\"F-*$)9$#\"\"$\"\"#F-!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 14 "c := n -> 1/n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"cGR6#%\"nG6\"6$%)operatorG%&arrowGF(*&\"\"\"F-9$!\"\"F(F(F(" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "Lets loo
k at what is happening graphically. The red graph is subject ak, while
 bk is blue and ck is green." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plot( [a(floor(x)), b(floor(
x)), c(floor(x)) ], x = 2..5, color = [red, blue,green]);" }}{PARA 13 
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}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "From th
is graph it appears that ak is larger than both bk and ck. And since  \+
ck diverges,  ak might diverge also!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "plot( [a(floor(x)), b(floo
r(x)), c(floor(x)) ], x = 5..20, color = [red, blue,green]);" }}{PARA 
13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6'-%'CURVESG6$7ax7$$\"\"
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cmFjco7$F`cmFjco7$FccmFjco7$FfcmFjco7$FicmFjco7$F\\dmFjco7$F_dmFjco7$F
bdmFjco7$FedmFjco7$FhdmFjco7$F[emFjco7$F^emFjco7$Faem$\"3G+++++++]F^s-
Ffem6&FhemF\\fmFiemF\\fm-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6$;F(Faem%
(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Cu
rve 1" "Curve 2" "Curve 3" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 63 "That theory was premature. We now see that thin
gs are changing." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 199 "Ultimately, ak and bk are more or less in sync. We can s
ee this we apply the Limit Comparison test and take the limits of the \+
terms. This does prove the limit comparison test, but makes it plausib
le." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "limit( a(n)/b(n), n = infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "li
mit( b(n)/c(n), n = infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"
!" }}}{PARA 0 "" 0 "" {TEXT -1 83 "___________________________________
________________________________________________" }}{PARA 4 "" 0 "" 
{TEXT -1 17 "C. The Ratio Test" }}{PARA 0 "" 0 "" {TEXT -1 83 "_______
______________________________________________________________________
______" }}{PARA 0 "" 0 "" {TEXT -1 220 "More generally, the ratio of c
onsecutive terms is an expression. The ratio test requires us to take \+
the limit of the absolute value of this ratio. When this limit is stri
ctly less than 1, the series converges absolutely." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Another important test i
s the Ratio test. In this test, we take the limit of the absolute valu
e of consecutive terms. If" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 45 "Sum( a*r^n, n = 0..infinity):   % = value(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\"nGF)/F
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"" {MPLTEXT 1 0 18 "a := n -> (6/7)^n;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"aGR6#%\"nG6\"6$%)operatorG%&arrowGF()#\"\"'\"\"(9$F(F(F(" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a(n+1)/a(n);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#*&)#\"\"'\"\"(,&%\"nG\"\"\"F*F*F*)F%F)!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6##\"\"'\"\"(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
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"" {XPPMATH 20 "6#/-%$SumG6$)#\"\"'\"\"(%\"nG/F+;\"\"!%)infinityGF*" }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 197 "Here w
e create a generic geometric series, then look at an example for an = \+
(6/7)^n. We take the ratio of consecutive terms, which is a constant o
f course. We then verify that the series converges." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a := n -> \+
(2^n) / n!;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aGR6#%\"nG6\"6$%)op
eratorG%&arrowGF(*&)\"\"#9$\"\"\"-%*factorialG6#F/!\"\"F(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a(n+1)/a(n);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#*&*&)\"\"#,&%\"nG\"\"\"F)F)F)-%*factorialG6#F(F)F)
*&-F+6#F'F))F&F(F)!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "
abs( simplify(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"\"F%-%$a
bsG6#,&%\"nGF%F%F%!\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 29 "R := limit( %, n = infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"RG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Sum( a(n), \+
n = 0..infinity): % = value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%
$SumG6$*&)\"\"#%\"nG\"\"\"-%*factorialG6#F*!\"\"/F*;\"\"!%)infinityG-%
$expG6#F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 58 "If the limit of the ratio is 1, the test is inconclusive. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "The ratio
 limit is 1 in this case, and the series diverges. However this is not
 always the case." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 20 "a := n -> 1/sqrt(n);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"aGR6#%\"nG6\"6$%)operatorG%&arrowGF(*&\"\"\"F--%%sq
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11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
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