# Finding all possible values of a so that the Taylor series expansion is well-defined

I am dealing with $$f(x) = \log_3(2x-1)$$ and I already found out the Taylor series expansion at $$a = 1$$ as $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2)^n(x-1)^n}{n\log(3)}$$. My problem is I need to find all possible values of a so that Taylor series expansion at a is well-defined. What does it mean if the Taylor series expansion is well-defined? Do I need to solve for the internal of convergence?

• I guess you need to check if radius of convergence $R>0$. I suppose that Taylor series expansion of $e^{-1/x^2}$ which is $0$ at $a=0$ can also be considered as not well defined even though its radius of convergence is obviously infinite, but we are not in this situation here.
– zwim
Jun 10 at 9:47

Using, as you did, natural logarithms and computing the successive derivatives, around $$x=a$$, you have $$\log_e(3)\,f(x)=\log(2a-1)-\sum_{n=1}^\infty \frac{\left(\frac{1}{2}-a\right)^{-n}}{n} (x-a)^n$$ So, the logarithm gives a first condition.
Now, for the radius of convergence $$b_n=\frac{\left(\frac{1}{2}-a\right)^{-n}}{n}\implies \frac{b_n}{b_{n+1}}=\frac{ n+1}{n}\left(\frac{1}{2}-a\right)\quad \to \left(\frac{1}{2}-a\right)$$ I am sure that you can conclude.