Convergence and Boundary Points in Power Series

[buffordboy23]

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

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}

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

\frac{1}{\left(e-1\right)}

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.

 

[Dick]

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.

 

[buffordboy23]

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

\left|a-1\right| < R

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.

 

[Dick]

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).