Transcript nth term test - Santa Cruz High School

Tests for Convergence and
Divergence
When
Always
check
To a n
lim
If lim a n  0
n 
Then the series
diverges
nth term test
n 
first
Use
 
n n
   1 
1  



n 1 n1  en
Examples:
Diverges
because
This does not
work in
reverse! Just
because
lim a n  0
Warning!
n 
does not
mean that the
series
converges.
1




lim
lim1  
 e0 0
 e n 
n

n
n
n
Geometric
series test
In a  ar  ar2  ar3  ...
If r  1 Then the series
converges, otherwise the series
diverges.
Only use
for
geometric
series…
duh

n
 2  converges
5 

n 1  3 
Make sure that
r 1
before using
n 1
2
n
a
converges
(

1
)
formula

n
to
3
n 1
1 r

 4 8  16
4
 
 ...
a

3 9 27
3
4
2
a
4
3
r



3
1 r 1  2
5
3
Telescoping Series




This is used to find the sum of the series,
not necessarily if it converges or diverges.
You might need to use partial fractions.
Write out the first few terms to see what
“telescopes” out.
Find the limit of Sn as n  ∞
f(x) is continuous, positive,and decreasingfor x
let a n  f (n)  f ( x)
Rarely
used
except
to
prove
P-test
a

n
and  f(x)dx both
1
converge or
1
The value of the
Integral is not
Necessarily
the Value
Warning!
to which
The series
converges
Integral Test
diverge

1

1 to the
Convert
dx 

2
2

1 x  1integral
improper
n 1 n  1
Example:
 lim arctan( x)]  lim(arctan b  arctan 1) 
b 
b
1
b 

2


4

Usually only
used
in
comparison
tests
1
T hep  series  p
n 1 n
Converges if p > 1
Diverges if 0 < p < 1

P-test

Only use
on series
in the form:

n 1
1
diverges with p  1


n 1 n
1
convergeswit h p  2

2
n 1 n

1
1
diverges with p 

2
n
n 1
1
p
n
Comparison
Test
c
converges  dn diverges
if an  cn for all n  N
n
series converges
if an  dn for all n  N
Use primarily for

sin n
n2
types of series
series diverges
if you have an an that is
less than a divergent or
more than a convergent
the comparison test is
inconclusive
When to Use:
Examples:

1
1
1

for
all
n
n
n

n
2 1 2
n 1 2  1
1
since  n converges by geometric series
2
Therefore series converges
Given thata n  0, and bn  0
 an 
  L where L is
If lim 
n    bn 


both finite and positive,
Limit a n and  bn
Then theseries
use when
you have
an effective
Comparison
Test
both
convergeor both
diverge
power in the

Therefore
the series converges
n

denominator
1 converges by
 
because
2
n
1
n13 / 2
P-test nn 1
n
3
1
2
2 n 2
2
n
n
lim n  1  lim 2
 lim 2
1
1
n 
n  n  1
n  n  1
n
3
2
Absolute Convergence Theorem
If  a n converges,then  an convergesabsolutely
If  an divergesbut  an converges,
then  an convergesconditiona
lly
Ratio Test
Good for mix
of factorials,
exponentials,
powers

2n

n 1 n!
an 1
  lim
n  a
n
If ρ < 1, series converges
If ρ > 1, series diverges
If ρ = 1, inconclusive
The series does not necessarily
converge to ρ!!!
If ρ = 1, you must use a
different test
2 n 1
n 1
2
 n!
2
(n  1)!
  lim
 lim
 lim
0
n
n
n 
n  ( n  1)!2
n  n  1
2
n!
  0  1 Therefore series converges
Root Test
Good when you
have something
complicated to
the nth power
2 
 

n 1  n! 

n
n
  lim n an
n
If ρ < 1, series converges
If ρ > 1, series diverges
If ρ = 1, inconclusive
The series does
not necessarily
converge to ρ
n
2 
2n
0 1
  lim n    lim
n


n!
n 
 n! 
n
Therefore series converges
Alternating
Series Test
Chaaaarlie let’s go to Candy
Mountain!