Asymptotics for the number of digits of the ratio of binomial coefficients 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). 
 A: 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.
A: This is only a partial solution.
Write $\log(a_{n}^n)=n\log(a_{n})+\sum_{i=1}^{n-1}\log(1+\frac{i}{a_{n}})$.
For the first term, we have $\log(a_{n})=\log(a)+\log(n)+o(1)$.
For the second term, we have the following
$$a_{n}\int_{0}^{\frac{n-1}{a_{n}}}\log(1+x)dx\leq \sum_{i=1}^{n-1}\log(1+\frac{i}{a_{n}})\leq a_{n}\int_{\frac{1}{a_{n}}}^{\frac{n}{a_{n}}}\log(1+x)dx$$
Computations shows that both sides are equivalent to $$n(a+1)(\log(1+\frac{1}{a})-1)$$
Hence,    $\log(a_{n}^n)=n\log(n) +n\left(\log(a)+(a+1)(\log(1+\frac{1}{a})-1)\right)$. Idem for $b_{n}^{n}$
So that $\frac{1}{n}\log(a_{n}^n)$ does not converge...
However, $\frac{1}{n}(\log(a_{n}^n)-\log(b_{n}^n))$ converges to 
$$\tilde{f}(a,b)= \log(\frac{a}{b})+b-a+(a+1)\log(1+\frac{1}{a})-(b+1)\log(1+\frac{1}{b})$$
