(Select the choice which gives complete correct information.) ee xk The series Ek=1 2k(1 - x)ka)...

Question

(Select the choice which gives complete correct information.) ee xk The series E
k=1 2k(1 - x)k
a) converges at x 0, only b) converges for all x c) has radius of convergence R =4 d) converges for x in half open interval (-2/3,2/3] only e) converges for all x
16. (Select the choice which gives complete. correct information.) The series z x - k=0 2
a) converges at x - 3, only b) converges for all x c) converges for 1
17. (Select the choice which gives complete correct information.) If Z akxk has radius of convergence 3, and Z bk xk has radius of k=0 k=0 convergence 5, then Z (ak 4 bk)xk k=0
a) has radius of convergence 5 b) has radius of convergence 8 c) has radius of convergence 2 d) has radius of convergence 3 e) has radius of convergence 15
18. (Select the choice which gives complete correct information.) go Assume that Z akxk converges when x = 9 and diverges when x = -12. k=0 What, if anything, can we say about what happens at x = -7?
a) converges conditionally but not absolutely b) converges absolutely c) diverges d) not possible to decide e) converges conditionally at x = -7 and absolutely for -7

005_5k-wcccq4e3.jpg

David · Accepted Answer

For each of the following problems, we wish to select the choice which gives complete, correct 
information.
15. The series  
 
 
 1 2 1
k
kk
k
x
S x
x





a) converges at 0x   only. 
b) converges for all x. 
c) has radius of convergence 2
3
R  . 
d) converges for x in the half-open interval  2 23 3,  only. 
e) converges for all x in    23, 2, .    
 
First we consider 1,x   for which the series alternates. In this case,  S x  converges if 
and only if the terms approach zero as k goes to infinity. We have
 
   
.
2 12 1
k
k
k kk
x x
a
xx
 
   
  

Thus we see that lim 0k
k
a

  if and only if
 
 
1,
2 1
x
x



which holds if and only if
  2 1 2 1 2 2,x x x x x        
 
which in turn holds if and only if
 0 2,x 
i.e., if and only if 2.x   
 
Next we consider the case 0 1,x   for which each term ka  in the series is nonnegative. 
We still have
 
   2 12 1
k
k
k kk
x x
a
xx
 
   
  
.
By the root test, we see that  S x  converges if 
 
 
 
1,
2 1
x
x



which holds if and only if
  2 1 2 2 ,

(Select the choice which gives complete correct information.) ee xk The series E k=1 2k(1 - x)k a) converges at x 0, only b) converges for all x c) has radius of convergence R =4 d) converges for x in...

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