This document clearly describes, with detailed notes and examples, how to evaluate/solve the following:
~ Indeterminate forms
~ The Limit Comparison Test
~ The Integral Test
~ The Alternating Series Test
~ Truncation Errors
~ Conditionally Convergent Series
as taught by the University of...
}
THE DIVERGENCE
) TEST
! HAVE TO RE " / SE
(2) THE RATIO TEST "" ☐ " "" "" " " " """"" " ⇐
WORK !
(3) THE P SERIES TEST
-
Con own :)
(4) THE COMPARISON TEST ☐ Focus
☐ USE IT IN LIMIT
COMPARISON TEST &
D USE THIS PROOF TO STATE FIRST TEST ! ☐ LCT
!
RECALL : THE COMPARISON TEST : PROVED IN FIRST '
EAR + ABLE To EXAMPLES .
COMA RING TWO SERIES :
CONDITION :
THEOREM : LET :O fan £ bn FOR SUFFICIENTLY LARGE M THEN :
,
HIGHER LOWER
DEDUCE :
(1) IF SERIES bn CONVERGES (c) ,
THEN an CONVERGES (c) Too
.
0 SINCE ans bn
LOWER HIGHER
(2) IF SERIES An DIVERGES (d) ,
THEN ibn DIVERGES (D) TOO -
☐ SINCE an bn
11 FOR "
REMARK : WHEN STATED SUFFICIENTLY LARGE M IN MATHS IT MEANS THAT
,
( ABOVE)
THERE EXISTS SOME NUMBER IT E SUCH THAT THE CONDITION IS TRUE
condition 's TRUE : >
FOR ALL VALUES OF M T T
n
EQUA¥cee☐ → EXISTS !
HELPFUL COMPUTING
HÉPITAL'S
IN
'
•
REVISE ! L RULE :
Limits !
" "
* WITHOUT
(
WORD Limit :
1 THE LIMIT COMPARISON TEST : LCT HAVE COMPARISON TEST ( CT )
BEHAVIOUR SERIES :
° STATES : DUE TO OF
CONDITION ; CAN BE 1- VE FOR SUFFICIENTLY LARGE M !
( for my
(1) LET anATÑ BE POSITIVE TERMS AND
I. FROM SERIES HAVE TO IDENTIFY THE SERIES TERMS
an AND bn
CONDITION :
a" CONSTANT /
( 2)) IF TAKE LIMIT OF RATIO AS N > N IT RESULTS IN A NUMBER
bn ,
K to :
lim an
= K O
n > a bn
i. IF CONDITIONS ARE SATISFIED :
(3)THEN WHAT FOLLOWS IS THAT :
SERIES OF SERIES Ot -
Éln AND bin EITHER BOTH CONVERGE OR BOTH DIVERGE .
LOOK AT PROOF OF THIS TEST AND
EXAMPLES . . .
, PROOF OF LIMIT COMPARISON TEST :
(1) BY SUPPOSITION WE HAVE :
,
lim an
=
K WHERE 14=10 ; K' >0 SINCE 0
; bn >
n→x an > 0
bn
¥yg¥.ES :O LIMIT
IS K
(2) FOR SUFFICIENTLY LARGE N
,
WE HAVE
B%IÑ☐É☐ -
CHOOSE -10
An B°{Y☐mr
II -
bn £2k
in
K { Y§ BETWEEN : Fumi)
•
L *
' >
(LIMIT)
An
( CAN FIND AN INTEGER TEL SUCH THAT THE CONDITION : ÉK t bn £2K
IS TRUE FOR ALL NYT )
an
(3) IK -
bn f 2k IMPLIES THAT
an 1 tzkbn . . . . (1)
an £ Zkbn . . .
.
(2)
REMEMBER DO NOT KNOW
*
CONVERGENCE STATUS OF an AND bn !
(1)
(ASSUME)
(3) SUPPOSE Ian CONVERGES
,
THEN BY EQN :
( SERIES)
Zan
bn £
K
0 HIGHER THAN bn !
AND SO ⇐ an
CONVERGES
,
THEN BY COMPARISON TEST
,
{ bn MUST CONVERGE .
:c BOTH SERIES CONVERGE .
( ASSUME )
(4) SUPPOSE San DIVERGES THEN BY EQN (2) :
,
an £ Zkbn
M SMALLER THAN
an
•
f bn
2k
i.
AND so # An DIVERGES
,
THEN BY COMPARISON TEST bn MUST DIVERGE
, .
• : THEY BOTH DIVERGE .
REVERSE :
IF Ikan
'
A- CAN ALSO SAY FROM EQN (2) ÷ bn CONVERGES ÷ CONVERGES By C T
-
.
1
FROM EQN (1) 8 IF { bn DIVERGES %
2¥ DIVERGES BY C T
- . !
HENCE IN SUMMARY EITHER BOTH ,
an AND ibn CONVERGE
, ,
OR BOTH DIVERGE .
, HOW DO APPLY LCT ? PROCEDURE FOR COMPUTING LCT :
NOTE :
IF GIVEN SERIES An
( an bn )
EXPRESSION ASYMPTOTIC
I FIND bn SUCH THAT = bn HAS THE SAME BEHAVIOUR
AS An FOR LARGE n (
100k
Asn → D) AND CALL THAT EXPRESSION
:
N COMPUTE LIMIT
A"
2 FIND THE
L 'M : CHECK IF THIS LIMIT IS ( AND 1- VE)
n -sa
bn NON-ZER_O .
3 DETERMINE THE CONVERGENCE STATUS WHETHER THE SERIES CONVERGES
DIVERGES OF bn .
4 CONCLUDE THAT an CONVERGES DIVERGES , DEPENDING ON STATUS OF
CONVERGENCE bn .
ALSO NOTE THERE ARE SOME HINTS THAT IS USED .
, MEANING THAT
0 EG .
FOR N LARGE ( - : h SO
,
WE KNOW THAT
g
n
> .
'
.tn IS SMALL
:( CLOSE TO ZERO )
• USING TAYLOR EXPANSION
,
CAN PROVE THE FFG :
FOR SMALL ✗ ( MEANING ✗ IS APPROX
"imn%ÉÑÉ% WE HAVE THE FFG
APPROXIMATIONS VERY USEFUL IN LCT ! : ( For small x !)
I sink = x
2 tank xx
3 Cos>c = 1- Ex
arctank = X
4
'
(a)
-
tan =x
5 In / + x = x
☒ IF TAKE MACLAURIN SERIES EXPANSION OF :
✗
)
→ ' SA NUMBER
( / tx
USING BINOMIAL SERIES :
✗
6 ( It >c) = I + xx
EXAMPLES :
3h2 tzntl
(1) DETERMINE WHETHER THE SERIES n3+1
CONVERGES OR DIVERGES :
•
SERIES / SUMMATION TERM :
3h22 12h I •
1
Let an =
n
>
+I
2 FIND bn : SUCH THAT bn HAS THE SAME ASYMPTOTIC BEHAVIOUR AS an
i. FOR n LARGE (As n → a) %
Be IN THIS CASE : 0 WHEN COMPUTING LIMITS :
FUNCTION OF N POLYNOMIAL IN n
HAVE A RATIONAL POWER
POLYNOMIAL 'N n
1 IDENTIFY THE HIGHEST
} }
DENOMINATOR on
'
DIVIDE EACH TERM BY N !
3%+33+1
of n in
. .
,
An y
nn ? 1ns
+
I 3h +
÷z £3 +
9- + ¥3
> 0 :
Me KNOW THAT ALL TERMS n
BUT HAVE TO IDENTIFY A FUNCTION
WHOSE BEHAVIOUR REFLECTS
I 2 +
÷ % +
D
¥3 -1 AJAY
→
REMAINDER I + >
IF HAVE :
f. CHOOSE ANY
}
POLYNOMIAL CAN
TERM TO BE
POLYNOMIAL REMAINDER !
" TERMS
OR POWER
LOWER IN
TAKE HIGHEST POWER OF NUMARATOR AND HIGHEST IGNORE
A- TO USUALLY DETECT Behaviour of an :
POWER OF DENOMINATOR AND DIVIDE Two to NUMERATOR 1- DENOMINATOR
3
i. An =
no
3
i. Let bn
no
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