I need some peer review of this problem

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    Peer review Review
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Discussion Overview

The discussion revolves around determining the radius and interval of convergence for the series \(\sum_{n=1}^\infty\frac{(-1)^{(n+1)}nx^n}{2^n}\). Participants explore the correctness of proposed answers and the conditions for convergence at the endpoints of the interval.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims the radius is \(x < 2\) and the interval is \([-2, 2)\).
  • Another participant questions the convergence at \(x = -2\) and suggests the interval should be \((-2, 2)\).
  • A participant reports that their calculator indicates convergence at \([-2, 2)\).
  • Another participant advises checking convergence at \(x = -2\) using manual methods, suggesting reliance on calculators may be misleading.
  • A participant acknowledges a mistake in applying the alternating series convergence test and concludes that the series diverges at \(x = -2\).
  • One participant analyzes the general term at the endpoints \(x = 2\) and \(x = -2\) and states that the series does not converge at either endpoint because the terms do not approach zero.

Areas of Agreement / Disagreement

Participants express differing views on the interval of convergence, with some asserting it includes \(-2\) while others argue it does not. The discussion remains unresolved regarding the correct interval.

Contextual Notes

Participants reference different methods for testing convergence, including the alternating series test and the nth term test for divergence, indicating potential limitations in their approaches and assumptions about convergence behavior at the endpoints.

RadiationX
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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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[edit]
wait a minute
just noticed something
u sure it converges for -2?
the interval should (-2,2)
[/edit]
 
Last edited:
when i put this in my calculator it adds up at [-2,2).
 
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
 
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.
 
have i come to the correct conclusion shmoe?
 
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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