Find the derivative of a function that contains a sum If $$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$ find $f''(-2)$.

I know that, by ratio test, the previous sum converges iff $x\in(0,6)$.
I did:
$$\begin{matrix}
f'(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}\right)'&=&
\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}n{(x-3)}^{n-1}}&=&
\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}{(x-3)}^{n-1}}
\\
f''(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}{(x-3)}^{n-1}}\right)'&=&
\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}(n-1){(x-3)}^{n-2}},
\end{matrix}$$
so I did the radio test on the last sum:
$$\begin{matrix}
&&\displaystyle\lim_{n\to\infty}{\left|\dfrac{a_{n+1}}{a_n}\right|} \\
&=&\displaystyle\lim_{n\to\infty}{\left|\dfrac{{(-1)}^{n+2}n{(x-3)}^{n-1}}{3^{n+1}}\dfrac{3^n}{{(-1)}^{n+1}(n-1){(x-3)}^{n-2}}\right|}\\
&=&\dfrac{|x-3|}{3}\underbrace{\displaystyle\lim_{n\to\infty}{\left|\dfrac{n}{n-1}\right|}}_{=\:1}\\
&\Rightarrow&|x-3|<3\\
&\Rightarrow&x\in(0,6),&
\end{matrix}$$
and since $-2\not\in(0,6)$ we cannot find $f''(-2)$.

First question: is my reasoning right?
If yes, to my amazement the radius of convergence of both functions are the same. Out of curiosity, why is this happening? The convergence can be generalize to $n$-th derivative?
Thank you!
 A: Yes, it always works. It can be proved (this is a standard result about power series) that if the power series $\displaystyle\sum_{n=0}^\infty a_n(z-a)^n$ has radius of convergence $R$ then;


*

*the radius of convergence of the pwer series $\displaystyle\sum_{n=1}^\infty na_n(z-a)^{n-1}$ is also $R$;

*$\displaystyle\left(\sum_{n=0}^\infty a_n(z-a)^n\right)'=\sum_{n=1}^\infty na_n(z-a)^{n-1}$.

A: In terms of finding $f''(-2)$ with 
$$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$
then consider the following. 
Method 1
Since
\begin{align}
f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} = \ln\left(\frac{x}{3}\right)
\end{align}
then 
\begin{align}
f'(x) &= \frac{1}{x} \\
f''(x) &= - \frac{1}{x^2}
\end{align}
Method 2
\begin{align}
f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} \\
f'(x) &= \frac{1}{3} \, \sum_{n=1}^{\infty} \left( \frac{3 - x}{3} \right)^{n} = \frac{1}{3 \, \left(1 - \left( 1 - \frac{x}{3} \right) \right)} = \frac{1}{x} \\
f''(x) &= - \frac{1}{x^2}.
\end{align}
The value sought can now be obtained by setting $x = -2$.
