Convergence and Boundary Points in Power Series

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Homework Help Overview

The discussion revolves around the convergence and boundary points of a power series defined at the point a = 1. The original poster is grappling with the correct formulation of the power series and its implications for convergence, particularly when considering complex variables.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to address an issue with the n = 0 term in their power series and questions whether it can be adjusted without affecting convergence tests. They also explore the implications of boundary values in relation to the radius of convergence for complex variables.

Discussion Status

Some participants have offered guidance on simplifying the power series by moving terms outside the summation. Others have pointed out that the radius of convergence does not provide information about the behavior at boundary points, suggesting that these need to be evaluated separately.

Contextual Notes

The original poster is working within the constraints of a homework assignment, which may impose specific rules regarding the treatment of power series and convergence. There is also an underlying assumption about the nature of singularities and their relationship to convergence.

buffordboy23
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I have defined a power series about the point a = 1 as,

[tex]f\left(a\right)= \sum^{\infty}_{n=0}\left(-1\right)^{n}\left[-\frac{1}{\left(e-1\right)^{n}}+\frac{1}{\left(e-1\right)^{n+1}}\right]\left(a-1\right)^{n}[/tex]

The terms of the power series are correct for all n, except n = 0. I need n = 0 to be equivalent to

[tex]\frac{1}{\left(e-1\right)}[/tex]

which is not, since a -1 accompanies it. The only way I can think of rectifying this problem is pulling the term outside of the summation sign and adding it to the summation series over n = 1 to infinity. Is it possible to write this power series without doing this procedure? I tried different ideas but no luck. Also, if I do decide to pull out this term, would it later affect the results of a ratio test for convergence? Thanks in advance.
 
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Just put it outside the summation. Keep it simple. No finite number of terms can affect the result of any convergence test. You only care about the limit n -> infinity.
 
Thanks Dick.

Here is another question. Suppose "a" represents a complex variable. According to Taylor's theorem, the power series expansion of my original function (not shown) has a radius of convergence given by

[tex]\left|a-1\right| < R[/tex]

where R is the distance to the nearest singularity from the point a = 1--this will give me two boundary values of the complex variable "a" for which the series converges. If I establish convergence when both of the boundary values are placed into my power series of the last post and although one of the boundary values is the point of the singularity itself, does this imply that the radius of convergence then equals the radius R? The textbook did not make this distinction, and I wonder if a radius of convergence can include the point of singularity. Obviously, the series will be divergent for a > R.
 
Determining the radius of convergence is R tells you nothing about points on the boundary. You have to consider them separately. The classic example is the expansion of log(1+x). At x=-1 it's divergent (no surprise, that's a singularity). At x=1 it's an alternating series converging to log(2).
 

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