Alternating Series and Intervals of Convergence
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#1(2 pts) Find a power series with interval of convergence [ 1, 3) . No work necessary.
( x  1)n

n
n 1 n  2

#2(5 pts each) Determine whether the following series converge absolutely, converge
conditionally, or diverge. Show all work and explain all steps thoroughly.
(1)k 2k

k2
k 1
Look for Absolute Convrgence:

a.


k 1


(1) k 2k
2k
2


. This series converges by p – test with p = 3/2.


2
2
3/2
k
k 1 k
k 1 k
(1)k 2k
converges absolutely.

k2
k 1

Therefore,

b.
1
sin  
k
k 1
  1 
1
 1 
Since, sin    sin 
sin     0 ,
  0 for all k  1 and lim
k 0 
k
 k 1 
  k 

1
(1) k sin   converges by A.S.T.

k
k 1
 (1)
k
1
sin  
 n   lim sin  x   1
lim
n 
x 0
1
x
n


1
1 
1
Since,  diverges,  (1) k sin     sin   diverges by L.C.T..
 k  k 1  k 
k 1 k
k 1

Therefore
 (1)
k 1
k
1
sin   converges conditionally.
k

#3(2 pts each) Given that S ( x)   ak ( x  2) k converges for x  5 and diverges for x = 2 ,
k 1
determine whether each of the following statements must be true, may be true, or must be false.
Give a brief explanation:
a. The radius of convergence is S ( x ) is 5.
This must be false. 3  R  4 .
b. The series diverges at x  6 .
This may be true. The interval of convergence may be (6, 2) or [6, 2)
c. The interval of convergence for S(x) is [5,1] .
This may be true. The center of the interval of convergence must be at x  2 , and since
the series converges at 5, it may also converge at 1.
d. The series converges absolutely on the interval ( 5,1) .
This must be true. Since the series converges at x  5 , the interval of convergence
must include at least [ 5,1) ; the series must converge absolutely on the interior of this
interval.
ln(k )  ( x  3)k
. Show all

k 2  2k
k 1

#4(6 pts). Determine the interval of convergence of the power series
work and explain all steps thoroughly.
Absolute Ratio Test:
 ( x  3) k 1  ln(k  1)

 x  3 ln(k  1)
k 2  2k
k2  x 3
lim 


lim


L
 k  
2
k 1
k
2
k 
(
k

1)

2
(
x

3)

ln(
k
)
2

ln(
k
)
(
k

1)
2




x 3
 1, then 1  x  5
So if
2


ln(k )  ( x  3)k
ln(k )  (1) k
Endpoints: x  1  
which converges by A.S.T.


k 2  2k
k2
k 1
k 1

ln(k )  ( x  3)k
ln(k )
x 5
 2 .
2
k
k 2
k 1
k 1 k

Since 0 

ln(k )
k
1
1


and
converes,

2
2
3/2
3/2
k
k
k
k 1 k
Therefore the interval of convergence is [1,5]

ln( k )
converges by C.T.
2
k 1 k

(1)k 2k
#5(5 pts). The series 
converges. If S 8 is used to approximate the value of this
k2
k 1
series, will the error be less 0.01? Explain your analysis clearly.

Since this is an alternating series with
that S  S8  a9 
2(k  1)
2k
2k

 0 and 2  0 as k   , we know
2
2
k
(k  1)
k
29 3 2
2
2
 .01 , we cannot be certain that S 8 is within


. Since
2
27
(9)
81
27
.01 of S.
#6(3 pts) Find the set of all x 
(1) k  3k
converges.

k
2
k 1 x  k

such that the series
Let’s Use Absolute Ratio Test:
(1)k 1  3k 1 x k  k 2
3
k2
3
lim k 1


lim

 L
2
k
k
2
k  x
k  x  ( k  1)
 (k  1) (1)  3
x
3
L  1   1  x  3  x  3 or x  3 .
x
(1) k  3k
converges absolutely on (, 3) (3, ) and diverges on (3,3) .
k
2
k 1 x  k
We now check x  3, 3 .

So we know


(1)k  3k
1
which converges by p-test.


k
2
2
k 1 x  k
k 1 k


(1)k  3k
(1)k
x  3   k 2   2 which converges absolutely by p-test.
k 1 x  k
k 1 k

x  3  
Therefore, Convergence Set = (, 3] [3, )