What is wrong with this proof about real analytic functions? I will give a "proof" for the following result that seems to be incorrect. Can some one tell me what is wrong with my reasoning?
RESULT: Let $f$, $g$ and $h$ be three real analytic functions on $(a-R,a+R)$ with $R>0$ such that
$$f(x)=g(x)+h(x)$$ for all $x\in (a-R,a+R)$ . If  $R>0$ is the radius of convergence of the Taylor series of two of these functions, then the radius of convergence of the  Taylor series of the third function is also $R$.
PROOF: If I use the given equality between the functions, I get that for all $n$,
$$f^{(n)}(x)=g^{(n)}(x)+h^{(n)}(x),$$ and in particular,
$$f^{(n)}(a)=g^{(n)}(a)+h^{(n)}(a).$$
Let $A_{n}$, $B_{n}$ and $C_{n}$ be the general terms of the power series expansion of $f$, $g$ and $h$ respectively. I divide each term of the last inequality by $n!$, and so:
 $$
 A_{n}=\dfrac{f^{(n)}(a)}{n!}=\dfrac{g^{(n)}(a)}{n!}+\dfrac{h^{(n)}(a)}{n!}
 =B_{n}+C_{n}.\qquad(1)$$
Now I am going to use a theorem that says: the radius of convergence of a series $\sum\alpha_{n}(x-a)^{n}$ is $R$ if and only if for all $0<r<R$ there exists a positive constant $k$ such that we have for all $n$,
$$|\alpha_{n}|\leq \dfrac{k}{r^{n}}.$$
Let $0<r<R$, and suppose for example that the radius of convergence for the Taylor's series of $g$ and $h$ is $R$. By the cited theorem there exists positive constants $k_{1}$ and $k_{2}$ such that for all $n$,
$$|B_{n}|\leq \dfrac{k_{1}}{r^{n}}$$ and $$|C_{n}|\leq \dfrac{k_{2}}{r^{n}}.$$
Then according to (1) we have for all $n$,
$$|A_{n}|\leq |B_{n}+C_{n}|\leq \dfrac{(k_{1}+k_{2})}{r^{n}}.????!!!$$

Your approach  is   fine. We can proceed from 
\begin{align*}
|A_{n}|\leq |B_{n}+C_{n}|\leq \dfrac{k_{1}+k_{2}}{r^{n}}
\end{align*}
as  follows:

We  set  $k=k_1+k_2$ and obtain for $0<r<R$
\begin{align*}
|A_{n}|\leq \frac{k}{r^n} \qquad\text{for all } n
\end{align*}
  We  conclude the convergence radius of $f$ expanded at $0$ is at least $R$.

Note the phrase at least. We have shown the claim is valid for all $0<r<R$, but we didn't say anything about $r>R$. 
Let's consider for example $g(x)=\frac{1}{1-x}$ and $h(x)=-\frac{1}{1-x}+\frac{1}{1-\frac{x}{2}}$. Both function have convergence radius $R=1$ when expanded at $0$, but $f(x)=g(x)+h(x)=\frac{1}{1-\frac{x}{2}}$ has radius of convergence $R=2$.
