(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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(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

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Answered Same DayDec 21, 2021

Answer To: (Select the choice which gives complete correct information.) ee xk The series E k=1 2k(1 - x)k a)...

David answered on Dec 21 2021

121 Votes

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 ,x x x   

which in turn...

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(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...

Answer To: (Select the choice which gives complete correct information.) ee xk The series E k=1 2k(1 - x)k a)...

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