- #1
buffordboy23
- 548
- 2
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.
[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.