Ch.2 Limits and derivatives

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Transcript Ch.2 Limits and derivatives

The comparison tests
Theorem Suppose that  an and  bn are series with
positive terms, then
(i) If  bn is convergent and an  bn for all n, then  an is
also convergent.
(ii) If  bn is divergent and an  bn for all n, then  an is also
divergent.

1
 Ex. Determine whether  n
converges.
n 1 2  1
1
1
 n
 Sol.
So the series converges.
n
2 1 2

The limit comparison test
Theorem Suppose that  an and  bn are series with
positive terms. Suppose
an
lim
 c.
n  b
n
Then
(i) when c is a finite number and c>0, then either both series
converge or both diverge.
(ii) when c=0, then the convergence of  bn implies the
convergence of  an .
(iii) when c  , then the divergence of  bn implies the
divergence of  an .

Example


Ex. Determine whether the following series converges.



1
2n 2  3n
p 
(3) sin
(2) 2
(1)
n
n 1
n 1 ln ( n  1)
n 1
5  n5
Sol. (1) diverge. choose bn  1/ n1/ 2 then lim an / bn  2
n 
(2) diverge. take bn  1/ n
an / bn  
then lim
n 
(3) converge for p>1 and diverge for p  1 take bn  1/ n p
p
then lim an / bn  
n 
Question



Ex. Determine whether the series
converges or diverges.
Sol. an  a
ln
1
n
e
1
ln ln a
n
1
 ln a
n
diverge for 0  a  e
converge for a  e
a
n 1
ln
1
n
(a  0)
Alternating series

An alternating series is a series whose terms are
alternatively positive and negative. For example,

1 1 1
(1)n1
1     
2 3 4
n
n 1

The n-th term of an alternating series is of the form
an  (1) n 1 bn or an  (1) n bn
where bn is a positive number.
The alternating series test

Theorem If the alternating series

n 1
(

1)
bn  b1  b2  b3  b4  b5  b6 

n 1
(bn  0)
bn  0
satisfies (i) bn 1  bn for all n (ii) lim
n 
Then the alternating series is convergent.
(1)n1
Ex. The alternating harmonic series 
n
n 1
is convergent.


Example

Ex. Determine whether the following series converges.


(1)n1
(1)n1 n2
(1)
(  0)
(2) 3

n
n 1
n 1
n 1

Sol. (1) converge

(1)n1 n
Question. 
n 1 4n  1

(2) converge
Absolute convergence

A series  an is called absolutely convergent if the series
of absolute values | an | is convergent.
(1)n1
For example, the series  3/ 2 is absolutely convergent
n 1 n
while the alternating harmonic series is not.
A series  an is called conditionally convergent if it is
convergent but not absolutely convergent.
Theorem. If a series is absolutely convergent, then it is
convergent.




Example

Ex. Determine whether the following series is convergent.

sin n
(1) 2
n 1 n

(1)n
(2)
n 1 ln(1  n)

Sol. (1) absolutely convergent
(2) conditionally convergent
The ratio test

The ratio test

an 1
(1) If lim
 L  1, then  an is absolutely convergent.
n  a
n 1
n

an 1
an 1
(2) If lim
  then  an diverges.
 L  1 or lim
n  a
n  a
n 1
n
n
an1
(3) If lim
 1, the ratio test is inconclusive: that is, no
n  a
n

conclusion can be drawn about the convergence of
a
n 1
n
Example

Ex. Test the convergence of the series

an
(1)
n 1 n !


an n!
(2) n
n 1 n
Sol. (1) convergent
(2) convergent for a  e; divergent for a  e
an 1 a n 1 (n  1)! a n n !
a
a

/ n 

n 1
n
an
(n  1)
n
(1  1/ n)
e
a  e  an 1  an  lim an  0
n 
The root test

The root test

(1) If lim n | an |  L  1, then  an is absolutely convergent.
n 
n 1

(2) If lim n | an |  L  1 or lim n | an |   then  an diverges.
n 
n 
n 1
(3) If lim n | an |  1, the root test is inconclusive.
n 
Example


Ex. Test the convergence of the series

n 1
n
1

a  
n

n
n
1
an  lim

Sol. lim
n 
n 
1 a
a
n
convergent for a  1; divergent for 0  a  1
n

n
n
a  1  an 
 (n  )
1 n
(1  )
n
(a  0)
Rearrangements




If we rearrange the order of the term in a finite sum, then
of course the value of the sum remains unchanged. But this
is not the case for an infinite series.
By a rearrangement of an infinite series  an we mean a
series obtained by simply changing the order of the terms.
It turns out that: if  an is an absolutely convergent series
with sum s , then any rearrangement of  an has the
same sum s .
However, any conditionally convergent series can be
rearranged to give a different sum.
Example
Ex. Consider the alternating harmonic series
1 1 1 1 1
1        ln 2.
2 3 4 5 6
Multiplying this series by 1/ 2, we get

1 1 1 1 1 1
1
       ln 2.
2 4 6 8 10 12
2
1
1
1
1
1
or
0   0   0   0    ln 2.
2
4
6
8
2
Adding these two series, we obtain
1 1 1 1 1
3
1        ln 2.
3 2 5 7 4
2
Strategy for testing series



If we can see at a glance that lim an  0 then divergence
n 
n 1
1
an 
 0
2n  1
2
If a series is similar to a p-series, such as an algebraic form,
or a form containing factorial, then use comparison test.
n3  1
1
an  3
~ 3/ 2
2
3n  4n  2 3n
For an alternating series, use alternating series test.
3
n
an  (1)n 4
n 1
Strategy for testing series


If n-th powers appear in the series, use root test.
 n2
an  ne
If an  f (n), f decreasing and positive, use integral test.
1
an 
n(ln n)(ln ln n)
1
(1)n ln n
2n n !
1
(1) tan
(2)
(3)
(4)
n
(n  2)!
(ln n)ln n
n

Sol. (1) diverge (2) converge (3) diverge (4) converge
Power series

A power series is a series of the form

n
2
3
c
x

c

c
x

c
x

c
x

n
0
1
2
3
n 0
where x is a variable and cn are constants called coefficients
of series.
 For each fixed x, the power series is a usual series. We can
test for convergence or divergence.
 A power series may converge for some values of x and
diverge for other values of x. So the sum of the series is a
function s ( x )  c  c x  c x 2  c x 3 
0
1
2
3
Power series

For example, the power series

n
2
3
x

1

x

x

x


n 0
1
converges to s ( x) 
when 1  x  1.
1 x

More generally, A series of the form

n
2
c
(
x

a
)

c

c
(
x

a
)

c
(
x

a
)

n
0
1
2
n 0
is called a power series in (x-a) or a power series centered
at a or a power series about a.
Example



Ex. For what values of x is the power series
convergent?
Sol. By ratio test,
n
n
!
x

n 0
an 1
(n  1)! x n 1
lim
 lim
 lim(n  1) | x |
n
n  a
n 
n 
n! x
n
the power series diverges for all x  0, and only converges
when x=0.
Homework 24

Section 11.4: 24, 31, 32, 42, 46

Section 11.5: 14, 34

Section 11.6: 5, 13, 23

Section 11.7: 7, 8, 10, 15, 36