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Let $a$ and $b$ be distinct positive real numbers. Let $(a_n)$ and $(b_n)$ be sequences of natural numbers such that $a_n\sim an$ and $b_n\sim bn$. All the limit relations here are for $n\to\infty$. Let $p_n$ and $q_n$ be the coprime natural numbers such \begin{equation*} \frac{p_n}{q_n}=\frac{a_n^{(n)}}{b_n^{(n)}}\left[=\binom{a_n+n-1}n\Big/\binom{b_n+n-1}n\right], \end{equation*} where $x^{(n)}:=x(x+1)\dots(x+n-1)$ is the Pochhammer rising factorial.

It then appears that \begin{equation} \tfrac1n\,\ln p_n\to f(a,b)\tag{1} \end{equation} for some positive real function $f$ and any distinct positive real numbers $a$ and $b$. (Here, in view of an immediate cancellation, such as $\frac{2\cdot3\cdot4}{3\cdot4\cdot5}=\frac25$, without loss of generality $a+1\le b$.)
This may be not hard to prove, using the prime number theorem (or maybe some refinement of it) together maybe with summation by parts.

Some easy remarks added: By Stirling's formula, $\tfrac1n\,\ln a_n^{(n)}\sim\ln n\to\infty$. However, because of the cancellations, $\frac1n\,\ln p_n$ seems likely to have a finite limit. Anyway, $\limsup_n \frac1n\,\ln p_n\le(a+1)\ln 2<\infty$, since $p_n\le\binom{a_n+n-1}n\le2^{a_n+n-1}=2^{(a+1+o(1))n}$.

However, $(1)$ may be well known. In such a case, it would be good to have a reference. Otherwise, it would be good to have a hopefully short and efficient proof, preferably with an explicit expression for $f(a,b)$. The case when $a$ and $b$ are natural numbers would be enough for my current needs (which arose in some work in approximation theory/numerical analysis).

Let $a$ and $b$ be distinct positive real numbers. Let $(a_n)$ and $(b_n)$ be sequences of natural numbers such that $a_n\sim an$ and $b_n\sim bn$. All the limit relations here are for $n\to\infty$. Let $p_n$ and $q_n$ be the coprime natural numbers such \begin{equation*} \frac{p_n}{q_n}=\frac{a_n^{(n)}}{b_n^{(n)}}\left[=\binom{a_n+n-1}n\Big/\binom{b_n+n-1}n\right], \end{equation*} where $x^{(n)}:=x(x+1)\dots(x+n-1)$ is the Pochhammer rising factorial.

It then appears that \begin{equation} \tfrac1n\,\ln p_n\to f(a,b)\tag{1} \end{equation} for some positive real function $f$ and any distinct positive real numbers $a$ and $b$. (Here, in view of an immediate cancellation, such as $\frac{2\cdot3\cdot4}{3\cdot4\cdot5}=\frac25$, without loss of generality $a+1\le b$.)
This may be not hard to prove, using the prime number theorem (or maybe some refinement of it) together maybe with summation by parts.

Some easy remarks added: By Stirling's formula, $\tfrac1n\,\ln a_n^{(n)}\sim\ln n\to\infty$. However, because of the cancellations, $\frac1n\,\ln p_n$ seems likely to have a finite limit. Anyway, $\limsup_n \frac1n\,\ln p_n\le(a+1)\ln 2<\infty$, since $p_n\le\binom{a_n+n-1}n\le2^{a_n+n-1}=2^{(a+1+o(1))n}$.

However, $(1)$ may be well known. In such a case, it would be good to have a reference. Otherwise, it would be good to have a hopefully short and efficient proof, preferably with an explicit expression for $f(a,b)$. The case when $a$ and $b$ are natural numbers would be enough for my current needs (which arose in some work in approximation theory/numerical analysis).

Let $a$ and $b$ be distinct positive real numbers. Let $(a_n)$ and $(b_n)$ be sequences of natural numbers such that $a_n\sim an$ and $b_n\sim bn$. All the limit relations here are for $n\to\infty$. Let $p_n$ and $q_n$ be the coprime natural numbers such \begin{equation*} \frac{p_n}{q_n}=\frac{a_n^{(n)}}{b_n^{(n)}}\left[=\binom{a_n+n-1}n\Big/\binom{b_n+n-1}n\right], \end{equation*} where $x^{(n)}:=x(x+1)\dots(x+n-1)$ is the Pochhammer rising factorial.

It then appears that \begin{equation} \tfrac1n\,\ln p_n\to f(a,b)\tag{1} \end{equation} for some positive real function $f$ and any distinct positive real numbers $a$ and $b$. (Here, in view of an immediate cancellation, such as $\frac{2\cdot3\cdot4}{3\cdot4\cdot5}=\frac25$, without loss of generality $a+1\le b$.)
This may be not hard to prove, using the prime number theorem (or maybe some refinement of it) together maybe with summation by parts.

Some easy remarks added: By Stirling's formula, $\tfrac1n\,\ln a_n^{(n)}\sim\ln n\to\infty$. However, because of the cancellations, $\frac1n\,\ln p_n$ seems likely to have a finite limit. Anyway, $\limsup_n \frac1n\,\ln p_n\le(a+1)\ln 2<\infty$, since $p_n\le\binom{a_n+n-1}n\le2^{a_n+n-1}=2^{(a+1+o(1))n}$.

However, $(1)$ may be well known. In such a case, it would be good to have a reference. Otherwise, it would be good to have a hopefully short and efficient proof, preferably with an explicit expression for $f(a,b)$. The case when $a$ and $b$ are natural numbers would be enough for my current needs (which arose in some work in approximation theory/numerical analysis).

Let $a$ and $b$ be distinct positive real numbers. Let $(a_n)$ and $(b_n)$ be sequences of natural numbers such that $a_n\sim an$ and $b_n\sim bn$. All the limit relations here are for $n\to\infty$. Let $p_n$ and $q_n$ be the coprime natural numbers such \begin{equation*} \frac{p_n}{q_n}=\frac{a_n^{(n)}}{b_n^{(n)}}\left[=\binom{a_n+n-1}n\Big/\binom{b_n+n-1}n\right], \end{equation*} where $x^{(n)}:=x(x+1)\dots(x+n-1)$ is the Pochhammer rising factorial.

It then appears that \begin{equation} \tfrac1n\,\ln p_n\to f(a,b)\tag{1} \end{equation} for some positive real function $f$ and any distinct positive real numbers $a$ and $b$. (Here, in view of an immediate cancellation, such as $\frac{2\cdot3\cdot4}{3\cdot4\cdot5}=\frac25$, without loss of generality $a+1\le b$.)
This may be not hard to prove, using the prime number theorem (or maybe some refinement of it) together maybe with summation by parts.

Some easy remarks added: By Stirling's formula, $\tfrac1n\,\ln a_n^{(n)}\sim\ln n\to\infty$. However, because of the cancellations, $\frac1n\,\ln p_n$ seems likely to have a finite limit. Anyway, $\limsup_n \frac1n\,\ln p_n\le(a+1)\ln 2<\infty$, since $p_n\le\binom{a_n+n-1}n\le2^{a_n+n-1}=2^{(a+1+o(1))n}$.

However, $(1)$ may be well known. In such a case, it would be good to have a reference. Otherwise, it would be good to have a hopefully short and efficient proof, preferably with an explicit expression for $f(a,b)$. The case when $a$ and $b$ are natural numbers would be enough for my current needs (which arose in some work in approximation theory/numerical analysis).

Let $a$ and $b$ be distinct positive real numbers. Let $(a_n)$ and $(b_n)$ be sequences of natural numbers such that $a_n\sim an$ and $b_n\sim bn$. All the limit relations here are for $n\to\infty$. Let $p_n$ and $q_n$ be the coprime natural numbers such \begin{equation*} \frac{p_n}{q_n}=\frac{a_n^{(n)}}{b_n^{(n)}}\left[=\binom{a_n+n-1}n\Big/\binom{b_n+n-1}n\right], \end{equation*} where $x^{(n)}:=x(x+1)\dots(x+n-1)$ is the Pochhammer rising factorial.

It then appears that \begin{equation} \tfrac1n\,\ln p_n\to f(a,b) \end{equation} for some positive real function $f$ and any distinct positive real numbers $a$ and $b$. (Here, in view of an immediate cancellation, such as $\frac{2\cdot3\cdot4}{3\cdot4\cdot5}=\frac25$, without loss of generality $a+1\le b$.)
This may be not hard to prove, using the prime number theorem (or maybe some refinement of it) together maybe with summation by parts.

However, this may be well known. In such a case, it would be good to have a reference. Otherwise, it would be good to have a hopefully short and efficient proof, preferably with an explicit expression for $f(a,b)$. The case when $a$ and $b$ are natural numbers would be enough for my current needs (which arose in some work in approximation theory/numerical analysis).

Let $a$ and $b$ be distinct positive real numbers. Let $(a_n)$ and $(b_n)$ be sequences of natural numbers such that $a_n\sim an$ and $b_n\sim bn$. All the limit relations here are for $n\to\infty$. Let $p_n$ and $q_n$ be the coprime natural numbers such \begin{equation*} \frac{p_n}{q_n}=\frac{a_n^{(n)}}{b_n^{(n)}}\left[=\binom{a_n+n-1}n\Big/\binom{b_n+n-1}n\right], \end{equation*} where $x^{(n)}:=x(x+1)\dots(x+n-1)$ is the Pochhammer rising factorial.

It then appears that \begin{equation} \tfrac1n\,\ln p_n\to f(a,b)\tag{1} \end{equation} for some positive real function $f$ and any distinct positive real numbers $a$ and $b$. (Here, in view of an immediate cancellation, such as $\frac{2\cdot3\cdot4}{3\cdot4\cdot5}=\frac25$, without loss of generality $a+1\le b$.)
This may be not hard to prove, using the prime number theorem (or maybe some refinement of it) together maybe with summation by parts.

Some easy remarks added: By Stirling's formula, $\tfrac1n\,\ln a_n^{(n)}\sim\ln n\to\infty$. However, because of the cancellations, $\frac1n\,\ln p_n$ seems likely to have a finite limit. Anyway, $\limsup_n \frac1n\,\ln p_n\le(a+1)\ln 2<\infty$, since $p_n\le\binom{a_n+n-1}n\le2^{a_n+n-1}=2^{(a+1+o(1))n}$.

However, $(1)$ may be well known. In such a case, it would be good to have a reference. Otherwise, it would be good to have a hopefully short and efficient proof, preferably with an explicit expression for $f(a,b)$. The case when $a$ and $b$ are natural numbers would be enough for my current needs (which arose in some work in approximation theory/numerical analysis).