MA1521 1516S1 Seq & Series Summary PDF

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AY2015-16S1 MA1521 SEQUENCES & SERIES GEOMETRIC SEQUENCES / SERIES TEST n r n  0 if r  1 ; r   if r > 1; lim r n d.n. if r n    k 0  k 1  r k   r k 1  r k m th n term Test Any lim u n  0 Series diverges n  lim u n  0  lim u n 1 / u n   L  1 ...

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AY2015-16S1 MA1521

SEQUENCES & SERIES

GEOMETRIC SEQUENCES / SERIES

TEST

n r n  0 if r  1 ; r   if r > 1; lim r n d.n.e. if r

n 



r



r

  r k 1 

k

k 0 

k 1



k

k m



m

m `1

r

m 2

nth term Test

Any

lim u n  0

Series diverges

n 

lim u n  0



 ...

limu n 1 / u n   L  1



 u

 u k1  or

k

k



 u

k

 u k 1   lim

k m

k 1

 uk 

lim u n  0

Convergent

 u

k

 u k 1 

k m

Root Test

No conclusion

lim u1n/ n  L  [0,1)

Convergent

USEFUL RESULTS FOR CONVERGENCE TESTS

lim u1n/ n  L  1 (including n 

)



limn

1/n

n 

1



n 

n

y  lim 1    e y n  n 

y n  L  0,   n  x n



CONVERGENCE TESTS for

u

n

where

un  0

y

conv

n

n

iff



x

n



u

n p

n

converges iff

n q



(p-series)

u

1

n

p

n

n

converges for any integers p, q

converges iff

Limit Compariso n Test

lim

n 

y

n

=0

xn

conv

x 

conv



y



y lim n   n  x n

x

n

conv

n

div

n





y n

NG Wee Seng Email: [email protected] Tel: 65164673

n

n

n

p 1

contains

2

n



un

,

No conclusion

n 

n 2

Divergent



lim

 n ! 3

polynomials or exponential terms, .e.g.

lim un1/ n  L  1

 n  1! (n  1) n!

contains

n2

n 

n 

un

factorials, polynomials or exponential terms, .e.g.

Divergent

)

limu n 1 / u n   L  1

k

N

N



 u

n 

(including

with

No conclusion

limu n 1 / u n   L  [0,1)

Ratio Test

n

i

n 

TELESCOPING SERIES

u

n

n 

k 0

Series of the form

TYPE OF SERIES 

 -1

1 provided r  1 1r

 rm  rk  r  r

CONCLUSION S

CONDITIONS

n

div

n

3

n

nn

,

Typically for series of the form p (n )

 q( n)

where p(n) and q( n) contain a combination of powers of n , exponential terms and logarithmic terms) Apply this test when both Ratio & Root Test s fail or are difficult to apply

AY2015-16S1 MA1521

SEQUENCES & SERIES



For testing

yn 

E.g.

P(n) Q ( n ) , if we let





n 5

leading term in P xn  leading term in Q Then

yn 

xn 

let

n2 n5

Case 2

L  0  series converges for all x  R = infinity L    series diverges for all x except when x = a 



Case 3

o

2n  3 ln n 2 n 1

 c x  a 

n

L  0, L   xa  R

Solve the inequality L < 1 to get R

such that

1 n

d dx 

  1 u

ALTERNATING SERIES TEST for



  1 

or

n

n

n 1

un



c x  a  n 0

n

n





nc x  a 





n

n

n 0

Series converges if both conditions (i) and (ii) hold: (i)

un

is decreasing, i.e.

( Let u ( x)  ux . u (ii) lim u  0 n 

un 1  u n

' (x )  0  un

for all

n 1

n 1

  c  x  a dx   c

n

n

Alternating Series Test

 c 0 , a constant.

n

Differentiation & Integration of power series

,



Case 1

o

n 0

n 5

n2 ln n 5 n5 2 n3 ln n

o

where the series



y lim n  1 n  x n

E.g.  If

2 n  n 1 n3 2 n 3

n 0

n

n

--- (I)

x  a n 1  C

--- (II)

n 1

n > some N Theorem. Both the ‘differentiated’ and ‘integrated’ series (in (I) & (II) resp.) have the same radius of convergence as the original series

is decreasing)

n

TAYLOR AND MACLAURIN SERIES POWER SERIES 

n  cn x  a

is a power series centered at (about)

x

f ( n )( a) x  an n! n 0





a . When x

n 0

is the Taylor series of

f

at x = a (provided it

exists) (when a = 0, series is also known as the Maclaurin series) Some Taylor series can be obtained from geometric series.

= a, the series = c 0 

Radius of convergence of

 c x  a 

n

n

1

is the real number, R such

n 0

that the series converges whenever whenever

x  a  R ; to find R,

u n  c n  x  a n . NG Wee Seng Email: [email protected] Tel: 65164673

xa  R

compute

and diverges

L  lim

n 

u n1 un

where

, r

2 ( x2 ) 5

1 r   1 1 1 1 1 Example:    1 2 x 1  2( x  2  2) 5  2( x  2) 5 1   2( 5x 2)



If





n0

f

 f (n ) (a ) ( a) .  x  an   cn  x  a n , then cn  n! n! n 0

(n )

...