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{SECT 0 {PARA 4 "" 0 "" {TEXT -1 33 "Module 10  : Serious About Series
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 3 "" 0 "" {TEXT -1 24 "1002 \+
: Convergence Tests" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" 
{TEXT -1 17 "O B J E C T I V E" }}{PARA 0 "" 0 "" {TEXT -1 130 "In thi
s project we will examine and use some of the various tests that can b
e used to determine if a series converges or diverges." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 10 "S E T U P " }}
{PARA 0 "" 0 "" {TEXT -1 252 "In this project we will use the followin
g command packages. Type and execute this line before begining the pro
ject below. If you re-enter the worksheet for this project, be sure to
 re-execute this statement before jumping to any point in the workshee
t." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "restart; with(plots):" }}{PARA 7 "" 1 "" {TEXT -1 50 
"Warning, the name changecoords has been redefined\n" }}}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 5 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 5 "" 0 "" {TEXT 
-1 83 "_______________________________________________________________
____________________" }}{PARA 4 "" 0 "" {TEXT -1 33 "A. The Integral T
est & \020P- Series" }}{PARA 0 "" 0 "" {TEXT -1 83 "__________________
_________________________________________________________________" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 500 "There are many occasions where a series is clo
sely related to an integral. In these cases,the integral will converge
 or diverge if and only if the corresponding sum converges or diverges
. This result is calle the Integral Test. Remember a sequence is  func
tion defined on the domain of natural numbers. If it makes sense to de
fine the same function for all positive real numbers, and that functio
in is integrable ( able to be intergrated) , then the integral test wi
ll apply. Lets look at an example." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 85 "Here is a function, the sequence it defin
es, 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(" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 30 "seq( evalf( f(k)), k = 1..12);" }}{PARA 12 "" 
1 "" {XPPMATH 20 "6.$\"\"\"\"\"!$\"+0R`NN!#5$\"+)*3]C>F($\"++++]7F($\"
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$\"+BA,TFF/$\"+@hi0CF/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "
plot( [f(x), f( floor(x)), f(floor(x+1))], x = 1..12, thickness = [4,2
,2], color = [blue, coral, red]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 
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ezFjam7$$\"3Q$3_ShIQ2\"F\\zFjam7$FjzFjam7$$\"3y;HKLDb!4\"F\\zFjam7$$\"
3GLe9VUk'4\"F\\zFjam7$$\"3]i:gqr;)4\"F\\zFjam7$$\"3s\"Hd!)4!p*4\"F\\zF
jam7$$\"3Lc^yh:X+6F\\zFe\\l7$$\"3%4-8b-875\"F\\zFe\\l7$$\"3b&)3C*[u>5
\"F\\zFe\\l7$$\"3)*\\(oH&ft-6F\\zFe\\l7$$\"3B3-)y!=y06F\\zFe\\l7$F_[lF
e\\l7$$\"3!**\\i&eYs>6F\\zFe\\l7$Fd[lFe\\l7$Fi[lFe\\l7$$\"3+++]`*y\\;
\"F\\zFe\\l7$F^\\lFe\\l7$$\"3?vo/%[u?=\"F\\zFe\\l7$$\"37]7.c'\\!)=\"F
\\zFe\\l7$$\"3gPM-Us.\">\"F\\zFe\\l7$$\"31Dc,G[-%>\"F\\zFe\\l7$$\"3q=<
,@'=b>\"F\\zFe\\l7$$\"3a7y+9C,(>\"F\\zFe\\l7$$\"3afe]5$fx>\"F\\zFe\\l7
$$\"3O1R+2i])>\"F\\zFe\\l7$$\"3=`>].JD*>\"F\\zFe\\l7$Fc\\l$\"3CeR<$HiM
8#Fer-Fh\\l6&Fj\\lF\\]lF[]lF[]lFcbm-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEW
G6$;F(Fc\\l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 569 "Note that the blue curve is be
tween the two step functions colored yellow and red. If curve encloses
 infinite area then so does the larger step function because at all ti
mes it is greater than or equal to the curve. On the other hand, if th
e curve encloses a finite area, then so does the smaller step function
. Simple enough. However, the upper step function is the lower one shi
fted one unit to the right. In other words, the upper one is the same \+
as the lower one plus one additional term of a1 = 1. Consequently, the
 series converge or diverge as the integral does." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "Sum( f(k), k
 = 1..infinity):   % = evalf(value(%));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%$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$\"+\\`P7;
!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 340 
"In this case, we can actually compute the integral and the sums. Note
 that all three converge, however they are not equal. The value of the
 integral is between the value of the sums from k = 1.. infinity and k
 = 2..infinity - just as we saw on the graph above. Thus the integral \+
tells us about convergence & divergence but not exact values." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 119 "One important result that follows directly fro
m the integral test, is that series of the form  1/k^p converge only f
or " }}{PARA 0 "" 0 "" {TEXT -1 30 "p > 1, and diverges otherwise." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
78 "assume( q > 0);   p := q + 1;   Int( 1/x^p, x = 1..infinity ):   %
 = value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG,&%#q|irG\"\"\"F
'F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$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$*&\"
\"\"F()%\"xG,&%#q|irGF(F(F(!\"\"/F*;F(%)infinityGF0" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "Note that integral is f
inite for p > 1, and finite for p =< 1." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 5 "" 0 "" {TEXT -1 0 "" }}
{PARA 5 "" 0 "" {TEXT -1 83 "_________________________________________
__________________________________________" }}{PARA 4 "" 0 "" {TEXT 
-1 28 "B. The Limit Comparison Test" }}{PARA 0 "" 0 "" {TEXT -1 83 "__
______________________________________________________________________
___________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 311 "Another important test is the \+
Limit Comparison Test. A simplified version of this test asserts that \+
if the ratio of the terms of two different sequences have a finite, no
n-zero limit, then both converge or both diverge. Lets examine why tes
t works from a geometric point of view. Consider these three sequences
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 " bk
 is known to converge while  ck is known to diverge. For now we are no
t sure if  ak converges or diverges." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "a := n -> (3 + sqrt(n))/(n
^2 - sqrt(n) + 5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aGR6#%\"nG6
\"6$%)operatorG%&arrowGF(*&,&\"\"$\"\"\"-%%sqrtG6#9$F/F/,(*$)F3\"\"#F/
F/F0!\"\"\"\"&F/F8F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 
"b := n -> n^(-3/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bGR6#%\"nG
6\"6$%)operatorG%&arrowGF(*&\"\"\"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 look at what is happening graphically. The red grap
h 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(flo
or(x)), b(floor(x)), c(floor(x)) ], x = 2..5, color = [red, blue,green
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{TEXT -1 113 "From this graph it appears that ak is larger than both b
k 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(
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0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "That theory was prem
ature. We now see that things are changing." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
199 "Ultimately, ak and bk are more or less in sync. We can see this w
e apply the Limit Comparison test and take the limits of the terms. Th
is does prove the limit comparison test, but makes it plausible." }}
{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 "limit( b(n)/c(n
), n = infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 5 "" 
0 "" {TEXT -1 0 "" }}{PARA 5 "" 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 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 220 "More generally, \+
the ratio of consecutive terms is an expression. The ratio test requir
es us to take the limit of the absolute value of this ratio. When this
 limit is strictly less than 1, the series converges absolutely." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Another \+
important test is the Ratio test. In this test, we take the limit of t
he absolute value 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$*&%\"aG
\"\"\")%\"rG%\"nGF)/F,;\"\"!%)infinityG,$*&F(F),&F+F)F)!\"\"F4F4" }}}
{EXCHG {PARA 0 "> " 0 "" {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 1 0 44 "Sum( a(n), n = 0..infinity):   % = value(%)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$)#\"\"'\"\"(%\"nG/F+;\"
\"!%)infinityGF*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 197 "Here we create a generic geometric series, then look at \+
an example for an = (6/7)^n. We take the ratio of consecutive terms, w
hich is a constant of 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$%)operatorG%&arrowGF(*&)\"\"#9$\"\"\"-%*facto
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+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#,$*&\"\"\"F%-%$absG6#,&%\"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\"\"\"-%*facto
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{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 th
e 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$%)
operatorG%&arrowGF(*&\"\"\"F--%%sqrtG6#9$!\"\"F(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 19 "abs( a(n+1)/a(n) );" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$absG6#*&*$-%%sqrtG6#%\"nG\"\"\"F,*$-F)6#,&F+F,F,F,F,
!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$absG6#*&*$-%%sqrtG6#%\"nG\"\"\"F,*$
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Sum ( a(n), n=1..infinity): \+
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