Mathematics II (MATH2011A) - ALGEBRA_Chapter_2 (Notes + Fully worked out examples)

CHAPTER 2 : CONVERGENCE OF SERIES :


REMINDER
INFIRSTYEARLE-ITTHE-t-FGTET.CI
• :




}
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