Mathreview10a.
Find the convergence set for the following power series:
Look at these with the ratio test

k 2 xk

k 1 k  1
an 1 1
=
R
an
an  (n2 )(n  2) 
(n  1)2
n2
an 1 
an 

 = 1 for n large The radius of
n 11
n  1 an1  n  1(n  1)2 
convergence, R is 1 and the domain of convergence is –1<x<1
(n  1) 2
n2
k 2 ( x  2)k
a

a


n 1
n
3n 1
3n
3k
k 1

an  (n2 )3n1 

 = 3 for n large The radius of
an1  3n (n  1)2 
convergence, R is 3 and the domain of convergence is –3<y<3 where y=x-2. The domain for x is then:
–1<x<5
an  ln(n  2) 
1
1
( x  15)k

an 1 
an 

 = 1 for n large The radius of
ln(n  2)
ln(n  1) an 1  ln(n  1) 
k 0 ln(k  1)

convergence, R is 1 and the domain of convergence is –1<y<1 where y=x-15. However, for this series
starting at k=0 we have a problem as ln(0+1) is ln(1) is zero and we start off with a zero in the denominator.
Thus for k=0 as the start the series diverges and makes little sense if k starts at one then the series
converges for x in the domain 14<x<16.
an  (2n)(n  1)(n  2) 
2(n  1)
2n
2k ( x  3) k

an 1 
an 

 = 1 for n large
(n  1)(n  2)
n(n  1) an 1  2(n  1)n(n  1) 
k 1 k ( k  1)

The radius of convergence, R is 1 and the domain of convergence is –3<y<3 where y=x-3. The domain for
x is then: 2<x<4
(2) n 1
2n
(2 x  3) k
an 1  n 1 an  n

4
4
4k
k 0

an  4n12n 

  4 / 2  2 for n large The radius of
an1  4n 2n1 
convergence, R is 2 and the domain of convergence is –2<y<2 where y=x + 3/2. The domain for x is then:
–(7/2)<x<(1/2)
an  (n  1) ln(n  2) 
n
(n  1)
(1)k kx k
let y = (-x) an 1 

an 

 = 1 for n
ln(n  2) an 1  n ln(n  3) 
ln(n  3)
k 1 ln( k  2)

large The radius of convergence, R is 1 and the domain of convergence is –1<y<1 where y=-x. The domain
for x is then:
-1< -x <1 -x < 1 or x> -1 and -1<-x or 1> x therefore –1<x< 1
1
1 an  3n1 
( x  2)2 k
a

a

let
y
=
(x+2)^2

n 1
n

 = 3 for n large The radius of
3n 1
3n an1  3n 
3k
k 0

convergence, R is 3 and the domain of convergence is –3<y<3 where y=(x+2)^2. The domain for x is then:
( x  2) 2  3, ( x  2)  3, x  3  2 and
x  32

 2 (3x)
k
3k
k 1
2n
2n 1
an 
an 1 
1
1
an
 1/ 2 for n large The radius of convergence, R is 1/2 and
an 1
the domain of convergence is –1/2<y<1/2 where y=(3x)^3. The domain for x is
then: (3x)
3
 1/ 2, x3 
1
1
and
,x 3
3
2(3)
3 2
Determine whether each series converges absolutely, converges conditionally, or diverges.
Here we want to look at the absolute value of the constants for absolute convergence
n
n 1
(1)k 1 k
an  2
an 1 
2
2
n 1
(n  1)  1
k 1 k  1
an1  (n  1)(n2  1) 

=
2
an
 ((n  1)  1)n 

 

1
1
n( n  )
n(1  2 )
2




(n  1)
n
n



=
 n((n  1)  1   n((n  1)  1  (n  1) 1  1 

 (n  1) 2 
(n  1)  
(n  1) 





1 
1 


 n(1  2 ) 
1  2 
n 

 n 
for n large this is 1 so this series does not absolutely converge.

1
1
1  1 

n(1  )(1  ) 1  1  
n
n
 n  n 
However the numbers an 
n
decrease in absolute value to zero. If the terms of an alternating
n 1
2
series decrease in absolute value to zero, then the series converges. This is a conditionally convergent
series.
k 1
n 1
n
(1) k
an 
an 1 
2n  1 an
2(n  1)  1
k 1 2k  1


for large n The numbers an 
an 1

1 
2



 (n  1)(2n  1) 
n = 1

 
 (2(n  1)  1)n   2  1 

(n  1) 

n
gives for large n (1/2) . If the terms of an alternating series
2n  1
decrease in absolute value to zero, then the series converges. This is not a conditionally convergent series.
This series diverges.
(n  1)2
1
1
(1)k 1 k 2
n 2 an1  (n  1)2 en  1
an1  n 1 an  n
  2 n1   (1  2 )  <1 So for n

k
e
e
an
n
e
e
k 1
 ne
 e

large the ratio yields a value less then 1 such that this series absolutely converges.
(1)k (1  k 2 )
1  (n  1)2
1  n 2 an1  (1  (n  1)2 )n3 
a


 1 for n large
a


n
n 1
3
2 
n3
an
k3
(n  1)3
k 1
 (n  1) (1  n ) 
1  n2 1
such the this series does not absolutely converge…goes as 1/k. However the numbers an 

n3
n

for n large decrease in absolute value to zero. If the terms of an alternating series decrease in absolute
value to zero, then the series converges. This is a conditionally convergent series.
2(n  1)
2n an 1  2(n  1)n !  1
(1)k 1 2k

an 
an 1 

  which tends to zero for n
n ! an
(n  1)!
k!
k 1
 2n(n  1)!  n

large , Therefore this series converges absolutely
n 1
(1)k 1 k
n
an 1 
an 

(n  2)(n  3)
(n  1)(n  2)
k 1 ( k  1)( k  2)
an 1  (n  1)(n  1)  (1  1/ n)(1  1/ n)

 1 for n large such the this series does not absolutely

an
1 3 / n
 n(n  3) 
n
converge…goes as 1/k. However the numbers an 
for n large decrease in absolute
(n  1)(n  2)

value to zero. If the terms of an alternating series decrease in absolute value to zero, then the series
converges. This is a conditionally convergent series.
an 1  ln(ln n) 
1
(1)k 1
1

an 1 
an 

  1 for n large such
ln(ln(n  1))
ln(ln(n)) an
k  2 ln(ln k )
 ln(ln(n  1)) 
1
the this series does not absolutely converg. However the numbers an 
for n large
ln(ln(n))

decrease in absolute value to zero. If the terms of an alternating series decrease in absolute value to zero,
then the series converges. This is a conditionally convergent series.
ln(n  1)
ln n an1  ln(n  1)n2 
(1)k 1 ln k
an  2
an 1 

 1 for n large such the this

2 
n
(n  1) 2
an
k2
k 1
 ln n(n  1) 
ln( n)
series does not absolutely converge. However the numbers an 
for n large decrease in absolute
n2

value to zero. If the terms of an alternating series decrease in absolute value to zero, then the series
converges. This is a conditionally convergent series.


(1)
k 1
an 1
an
1/ n 1
1/ k
k 1
1
k 
k
an 1
 (n  1)  n 

 
 n  1 
1/ n
1/ n
1
an  n  
n

 1 
 (n  1) 

 n 1 
1/ n 1
 1 


 n 1 
n

n
n
n

n 1
(n  1)
n
n
1/ n
1
an  n  
n
This looks like for n large this is greater than 1
n Increases for increasing n. Therefore this series diverges
Given the series:
Estimate the sum of the series by taking the sum of the first four terms. How accurate is this estimate?
(1)k 1 (1)41 (1)31 (1)21 (1)11 1 1 1 1
 2(4)2  2(3)2  2(2) 2  2(1) 2  6  4  2  =

2 k 2
2 2 2 1
2
2
2
2
k 1 2
1

 1 1

1  16  ...   4  32  ... = (3/4) + (1/32) + … The next terms in the series will be smaller than
4
1/32 the number obtained is 0.78125 This sum is accurate to two places.
How many terms of the series are necessary to estimate its sum with three-place accuracy? What is this
estimate?
Next set of terms is
1
1
 10 = (1/256)= 0.00390625 gives .78515625 constant to three places 6 terms.
8
2 2
This gives the estimate of the sum to three place accuracy.
Next set of terms is
1
1
 14 = 1/4096 = 0.000244140625 gives 0.785400390625 constant to 4 places 8
12
2
2
terms
Next set of terms is
1
1
 18 =1/65536= 0.0000152587890625 gives 0.7854156494140625 constant to 5
16
2
2
places ten terms
Next set of terms is
1
1
 22 = 1/1048576= 0.00000095367431640625 gives 0.78541660308837890625
20
2
2
(1)k 1

2 k 2
k 1 2

Use the generalized ratio test to find all numbers x for which the given series converges.
1
1 an  n  1 
xk
an 1 
an 
= 1 for n large The radius of convergence, R is 1 and


n 1
n an1  n 
k 1 k

the domain of convergence is –1<x<1

1 an  (n  1)n1 
1
x
1  x 
a

(

1)
(

1)
=
a



n
 =  for n
  
  n 1
n n an1  nn
(n  1)( n 1)
 k  k 1
 k 
k 1

large The radius of convergence, R is  and the domain of convergence is –  <x<  it does not matter

k
k
k 1
what x goes into the series eventually k will dominate making the partial series go toward zero.