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). 



Asymptotics for the number of digits of the ratio of binomial coefficients
So here is an answer, edited first from an unproved guess at the correct $f(a,b)$ and then a terse proof. The following version has a few more details added.


I claim that
$$
f(a,b)=\int\_0^{b+1} \left(\left\lfloor \frac{a+1}x\right\rfloor
-\left\lfloor\frac ax\right\rfloor
-\left\lfloor\frac{b+1}x\right\rfloor
+\left\lfloor\frac bx\right\rfloor\right)^+\,dx.
$$


This is derived from the exact expression:
$$
\log p\_n=\sum\_p\left(\sum\_{k=1}^\infty \left(\left\lfloor \frac{a\_n+n-1}{p^k}\right\rfloor
-\left\lfloor \frac{a\_n-1}{p^k}\right\rfloor-
\left\lfloor \frac{b\_n+n-1}{p^k}\right\rfloor
+\left\lfloor \frac{b\_n-1}{p^k}\right\rfloor\right)\log p
\right)^+
$$


Firstly from the exact expression, I claim that the contribution of those primes less than $n^{2/3}$ is at most $n^{2/3}\log n$, so that it suffices to consider only those primes in the range $[n^{2/3},(b+2)n]$. To see this, first notice that for each $p$ and $k$, the expression in the inner parentheses takes only values 0 and $\pm 1$. Also each $p$ only makes a contribution for powers up to $\log n/\log p$, so the maximum contribution from any $p$ is $O((\log n/\log p)\times \log p)=O(\log n)$. Hence the upper bound for the combined contribution for $p<n^{2/3}$. For larger primes, clearly only the first power matters. 


Let 
$$
S\_n=\sum\_{p<(b+2)n}\left(\left(\left\lfloor \frac{a\_n+n-1}{p}\right\rfloor
-\left\lfloor \frac{a\_n-1}{p}\right\rfloor-
\left\lfloor \frac{b\_n+n-1}{p}\right\rfloor
+\left\lfloor \frac{b\_n-1}{p}\right\rfloor\right)\log p
\right)^+,
$$
so that by the above, $\log p\_n=S\_n+o(n)$. 


Now for $\epsilon>0$, define three functions:
\begin{align\*}
\bar g(x)&=
\min\left(\max\left(\left\lfloor \frac{a+1+\epsilon}x\right\rfloor
-\left\lfloor\frac{a-\epsilon}x\right\rfloor
-\left\lfloor\frac{b+1-\epsilon}x\right\rfloor
+\left\lfloor\frac{b+\epsilon}x\right\rfloor,0\right),1\right)\\
g(x)&=
\min\left(\max\left(\left\lfloor \frac{a+1}x\right\rfloor
-\left\lfloor\frac{a}x\right\rfloor
-\left\lfloor\frac{b+1}x\right\rfloor
+\left\lfloor\frac{b}x\right\rfloor,0\right),1\right)\\
\underline g(x)&=
\min\left(\max\left(\left\lfloor \frac{a+1-\epsilon}x\right\rfloor
-\left\lfloor\frac{a+\epsilon}x\right\rfloor
-\left\lfloor\frac{b+1+\epsilon}x\right\rfloor
+\left\lfloor\frac{b-\epsilon}x\right\rfloor,0\right),1\right)\\
\end{align\*}


For all large $n$, we have $(a-\epsilon)n<a\_n<(a+\epsilon)n$ and $(b-\epsilon)<b\_n<(b+\epsilon)n$. For any such $n$, the $p$ summand in $S\_n$ is between $\underline g(p/n)$ and $\bar g(p/n)$.


I next claim that $\int\_0^{b+2} (\bar g-\underline g)=O(\epsilon\log(1/\epsilon))$. To see this, notice that if $x>2\epsilon$, $\lfloor \frac{c+\epsilon}x\rfloor$ and $\lfloor \frac{c-\epsilon}x\rfloor$ differ by at most 1, and they differ if $x\in \big((c-\epsilon)/n,(c+\epsilon)/n\big]$
for some $n$. The measure of the set where they differ is therefore $O(\epsilon\log(1/\epsilon))$, which proves the claim.


Now let $\underline h$ and $\bar h$ be continuous functions such that $\underline h\le \underline g\le \bar g\le\bar h$ and $\int\_0^{b+2}(\bar h-\underline h)
\le 2\int\_0^{b+2}(\bar g-\underline g)$.


For large $n$, we now have
$$
\sum\_{p<(b+2)n}\underline h(p/n)\log p \le S\_n
\le \sum\_{p<(b+2)n}\bar h(p/n)\log p.
$$


An equivalent formulation of the prime number theorem is that
$\frac 1n\sum\_{p<(b+2)n}\log p\,\delta\_{p/n}$ converges in the weak topology to
Lebesgue measure restricted to $[0,b+2]$. (This is a consequence of the fact that $\sum\_{p<n}\log p=n+o(n)$).


Hence $S\_n-n\int\_0^{b+2}g=o(n)$, as required.


