# Radius of convergence of $\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n$

I've been working on the following problem:

" Compute the Taylor series of the function $f(x)=\frac{1}{1-x}$ at $x=2$ and determine the radius of convergence $r$ "

I know that the given Taylor series is $\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n$, but I don't know how to determine the radius of convergence.

I have tried to find an expression for the Taylor series of $-\ln|1-x|$ at $x=2$ instead, because I was hoping to find a nice looking geometric series from which $r$ could easily be found. I have also tried to use the ratio test on $\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n$, but I don't know what to do with the derivatives.

• Well, have you tried computing the derivatives? – Eric Wofsey Jun 3 '18 at 20:50

Hint: (General.) Radius of convergence of power series $\sum_n a_n(x-x_0)^n$ is given by formula $$r = \frac 1{\limsup_n |a_n|^{1/n}}.$$
Hint:(Specific.) $$\frac 1{1-x} = \frac 1{-1-(x-2)} = -\frac 1{1+(x-2)}$$ and thus the Taylor series around $x_0 = 2$ can be obtained from geometric series. Radius of convergence is the same as well.
Edit: $$\frac 1{1-x} = \frac 1{-1-(x-2)} = -\frac 1{1+(x-2)} =-\sum_{n=0}^\infty (-(x-2))^n = \sum_{n=0}^\infty (-1)^{n+1}(x-2)^n$$
• I don't think I understand your last hint. The power series that conv. to $\frac{1}{1-x}=-\frac{1}{1+(x-2)}$ is $\sum x^n$ with $r=1$, but I don't see how the Taylor series around $x_0=2$ can be obtained from this. – Tom Jun 3 '18 at 22:01