TITLE: Limit of ${n^2 \choose n}/{n^2+n-1 \choose n}$
QUESTION [5 upvotes]: I am trying to work out 
$$\lim_{n \to \infty} \frac{{n^2 \choose n}}{{n^2+n-1 \choose n}}.$$
Numerically it appears to be approximately $0.377$.  Is it possible to get an exact answer?

REPLY [2 votes]: Let $f(n)=\binom {n^2}{n}/\binom {n^2+n-1}{n}.$ For $n>0$ we have $$ f(n)=\left(n!^{-1}\prod_{j=0}^{n-1}(n^2-j)\right)/\left(n!^{-1}\prod_{j=0}^{n-1}(n^2+n-1-j\right)=$$ $$ =\prod_{j=0}^{n-1}(n^2-j)/(n^2+n-1-j)=$$ $$=\prod_{j=0}^{n-1}\left(1+\frac {n-1}{n^2+n-1-j}\right)^{-1}.$$ Therefore for $n>1$ the value of $f(n)$ is between  $$\left(1+\frac {n-1}{n^2+n-1}\right)^{-n}$$ $$\text {and }\quad \left(1+\frac {n-1}{n^2}\right)^{-n}.$$ Since $(1+1/n)^{-n}=e^{-1}(1+O(n^{-2}))$ as $n\to \infty$, it is easily shown that the upper and lower bounds (above) for $f(n)$  both converge to $e^{-1}.$