Let $f$ be the function given by $$ f(x) = \frac{1}{x+1}. $$ As we easily check, the $k$-th derivative of $f$ is given by $$ f^{(k)}(x) = (-1)^k \frac{k!}{(x+1)^{k+1}}. $$ In particular $|f^{(k)}(x)| = \frac{k!}{(x+1)^{k+1}}$.
Suppose that $R_n$, $n \ge 1$ is a rational function such that
- $\displaystyle R_n(x) = \frac{P_n(x)}{Q_n(x)}$, where $P_n$ and $Q_n$ are polynomials,
- $R_n$ is proper, that is the degree of $Q_n$ is greater than the degree of $P_n$.
- all coefficients of $P_n$ and $Q_n$ are positive, (EDIT:) and $Q_n(0) \neq 0$,
and $R_n$ converges to $f$, in the sense that $$ \lim_{n \to \infty} R_n(x) = f(x). $$
Does it follow that the $k$-th derivative of $R_n$ is uniformly (in $k$ and $x$) bounded on the positive axis by the $k$-th derivative of $f$? In other words, is it true that there exists $C > 0$ such that $$ |R_n^{(k)}(x)| \le C \frac{k!}{(x+1)^{k+1}}, \qquad x > 0,\ k \ge 1? $$
Note that I would like to obtain the above estimate just for $x > 0$: in this interval $x \mapsto x + 1$ and $Q_n$ are positive (hence, nonzero).