Conjecture. Let $f$ be a real (or complex) analytic function defined on some open subset $U$ of real (complex) numbers and assume that $p,q,x\in U$ are such that $x\in B(p,r_p)\cap B(q, r_q)$, where $r_p$ and $r_q$ are radiuses of convergence of $f$ expanded at $p$ and $q$ respectively and where $B$ stands for an open ball. If $$\frac{|x-p|}{r_p}<\frac{|x-q|}{r_q},$$ then $P_n = o(Q_n)$, where $$P_n = \left|f(x) - \sum_{i=0}^n\frac{f^{(i)}(p)}{i!}(x-p)^i\right|\quad\text{and}\quad Q_n = \left|f(x) - \sum_{i=0}^n\frac{f^{(i)}(q)}{i!}(x-q)^i\right|$$
I came up with this conjecture while I was playing with an approximation of the natural logarithm using Taylor series. So, below I present the path of reasoning which lead me to this conjecture. I put a more detailed anaysis in this jupyter notebook.
If $p>0$ and $x\in (0, 2p)$, then $$\ln(x)=\ln(p) - \sum_{i=1}^\infty\frac{1}{i}\left(\frac{p-x}{p}\right)^i.$$
By looking at the above formula, we notice that the very first term is $\ln(p)$. Let us say that we want to approximate $\ln(x)$, so we forbid ourselves to use logarithm in this series. Hence, we constrain ourselves to $p$ of the form $e^K$, where $K$ can be any integer.
Problem. For a given point $x>0$, find $p=e^K$ such that Taylor series of $\ln$ expanded at $p$ converges to $\ln(x)$ in the fastest way.
We know that if $x$ close to an edge of the interval of convergence, then the Taylor series converges very slowly. Thus, using this basic fact, I found the answer heuristically. In a nutshell, the idea is to compare distance from $x$ to $p$ but also take into account the size of the interval of convergence. Thus, we just look for best $p$ such that $$\frac{|x-p|}{r_p}$$ is the smallest, where $r_p$ is the radius of the convergence of $\ln$ a $p$.
So, if $x\in (e^K, 2e^K)$, then we should compare $$\frac{x-e^K}{e^K}\quad\text{with}\quad\frac{e^{K+1}-x}{e^{K+1}}$$ and put $p = e^K$ if the first one is smaller and put $p = e^{K+1}$ otherwise.
For $x = C_0e^K$, where $C_0 = \frac{2e}{1+e}$, we have the equality of the above two numbers. Thus, in order to compute $\ln(x)$ using Taylor series, it is best to start with $p=e^K$, where $K$ is such that $x\in[C_0e^{K−1},C_0e^K)$.
I have done some numerical tests whether this $C_0$ is the optimal one and it looks like it is. You are welcome to see the details in this already mentioned jupyter notebook. So, I wonder if this property generalizes to an arbitrarty analytic function and the conjecture is a manifestation of my wonderings.