I need some peer review of this problem

In summary, the radius of convergence for the series \sum_{n=1}^\infty\frac{(-1)^{(n+1)}nx^n}{2^n} is x<2 and the interval of convergence is (-2,2). There was confusion about the interval being [-2,2) but it was clarified that the series does not converge at x=2 or x=-2. The conclusion is that the series diverges for both of these values.
  • #1
RadiationX
256
0
find the radius and interval of convergence of:

[tex]\sum_{n=1}^\infty\frac{(-1)^{(n+1)}nx^n}{2^n}[/tex]

the radius is [tex]x<2[/tex]

and the interval is [-2,2)

are these two answers correct?
 
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  • #2
Yes

-- AI
[edit]
wait a minute
just noticed something
u sure it converges for -2?
the interval should (-2,2)
[/edit]
 
Last edited:
  • #3
when i put this in my calculator it adds up at [-2,2).
 
  • #4
RadiationX said:
when i put this in my calculator it adds up at [-2,2).

Step 1) throw away your calculator

Step 2) check convergence at x= -2 with pencil and paper
 
  • #5
shmoe said:
Step 1) throw away your calculator

Step 2) check convergence at x= -2 with pencil and paper

ah i used the alternating series convergence test incorrectly. by the nth term test for divergence i get that the limit of the series is [tex]\infty[/tex] which is not equal to zero so it diverges.
 
  • #6
have i come to the correct conclusion shmoe?
 
  • #7
At x= 2, the general term is [tex](-1)^{n+1}n[/tex]
At x= -2, the general term is [tex](-1)^{2n+1}n[/tex]

The series does not converge at either of those because the term does not go to zero. The interval of convergence is (-2, 2).
 

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