What is the limit of $\dfrac{{{k-1}\choose n}}{{{n+k-1}\choose n}}$ as $k/n^2\to a>0$ as both $k,n\to\infty$?

Question

Suppose $n,k\to\infty$ such that $k/n^2\to a>0$. What is the limit of $\dfrac{{{k-1}\choose n}}{{{n+k-1}\choose n}}$?

Apparently there is a discrepancy between the correct answer $\exp(-1/a)$ and the answer I am getting after applying Stirling approximation. I wrote $$\dfrac{{{k-1}\choose n}}{{{n+k-1}\choose n}}=\dfrac{((k-1)!)^2}{(k-n-1)!(k+n-1)!}$$ Now applying Stirling formula to each factorial, taking $k$ and $n$ out wherever possible and using that $f(x)^{g(x)}\to \exp(\lim g(x)(f(x)-1))$, I find the limit is $\exp(-2/a)$ instead of $\exp(-1/a)$. Where am I going wrong?

Note we can apply Stirling because $k-n\to\infty$ since $k/n^2\to a$.

EDIT Let me write out my approach. I think I need to see why I went wrong. Let me replace $k-1$ by $k$ because $k/n^2\to a$ according to P. Quinton's suggestion in the comment. Then the quantity whose limit I wish to evaluate, is $\dfrac{(k!)^2}{(k-n)!(k+n)!}$.

We use the notation $f\sim g$ whenever $f/g\to1$. Applying Stirling formula, $$\dfrac{(k!)^2}{(k-n)!(k+n)!}\sim \dfrac{k^{2k+1}e^{-2k}}{(k-n)^{k-n+1/2}(k+n)^{k+n+1/2}e^{-2k}}$$ $$=\dfrac{k^{2k+1}}{k^{2k+1}(1-\frac{n}{k})^{k-n+1/2}(1+\frac{n}{k})^{k+n+1/2}}=(1-\frac{n}{k})^{n-k-1/2}(1+\frac{n}{k})^{-k-n-1/2}$$

Now I use $f(x)^{g(x)}\to e^{\lim g(x)(f(x)-1)}$ to get $$(1-\frac{n}{k})^{n-k-1/2}(1+\frac{n}{k})^{-k-n-1/2}\sim \exp(-\frac{n^2}{k}+n+\frac{n}{2k}-n-\frac{n^2}{k}-\frac{n}{2k})$$ $$=\exp(-\frac{2n^2}{k})\to e^{-\frac{2}{a}}$$

Answer 1

If you start letting $k=a n^2$ with $a >0$, the problem becomes simple just using Stirling approximations and continuing with Taylor series for large values of $n$.

In such a case (I hope that I am not totally wrong), you should get something like $$e^{-\frac 1a} \left( 1- \frac {3a+1 } { 6a^3n^2}+O \left(\frac 1{n^4}\right)\right)$$

Answer 2

It turns out that your error is revealed in your comment:

I am not sure if it is entirely correct, just something I knew as a tool... but $f(x)^{g(x)}=\exp(g(x)\log(1+f(x)−1))$ and if $f(x)−1$ goes to $0$ then this quantity is like $\exp(g(x)(f(x)−1))$ since $\log(1+x)∼x$.

Do not use asymptotic equivalents without understanding exactly what they mean. See this post for the correct way to use asymptotic expansions. Specifically, $\ln(1+x) ∈ x + O(x^2) ⊆ x + o(x)$ as $x→0$, so $\exp(g(x)·\ln(f(x))) ∈ \exp(g(x)·(f(x)-1+o(f(x)-1)))$ as $f(x)-1 → 0$, and you cannot ignore the little-o term because it is multiplied by $g(x)$.

Note that correct asymptotic analysis can never result in incorrect answers, and will allow you to automatically know when you need to refine your asymptotic expansions. Here, for example, we can see that we need to refine the bound on the $\ln(f(x))$ until the error term multiplied by $g(x)$ goes to zero.

Answer 3

Just a comment, after applying Stirling's approximation to

$$ y = \log \left(\frac{((k-1)!)^2}{(k-n-1)! (k-n+1)!}\right) , $$ I got $$ y = (-k+n+1) \log (k-n-1)+(-k+n-1) \log (k-n+1)+2 (k-1) \log (k-1)-2 n+2 . $$ After replacing $k$ by $a n^2$, I got $$ e^y = \frac{e^{2-2 n} (1-n (a n-1))^{-a n^2+n+1} (n (a n-1)+1)^{-a n^2+n-1} }{ \left(1-a n^2\right)^{2 -2 a n^2} }. $$ That is a $0/0$ situation. I applied a few L'Hoppitals and I gave up.