Limit of ${n^2 \choose n}/{n^2+n-1 \choose n}$ 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?
 A: We have that as $n\to+\infty$,
$$\ln\left(\frac{{n^2 \choose n}}{{n^2+n-1 \choose n}}\right)=
\ln\left(\frac{\prod_{k=1}^{n-1}\left(1-\frac{k}{n^2}\right)}{\prod_{k=1}^{n-1}\left(1+\frac{k}{n^2}\right)}\right)
=\sum_{k=1}^{n-1}\ln\left(1-\frac{k}{n^2}\right)-\sum_{k=1}^{n-1}\ln\left(1+\frac{k}{n^2}\right)\\
= -\frac{2}{n^2}\sum_{k=1}^{n-1}k+o(1)=-\frac{n(n-1)}{n^2}\to -1$$
where we used the fact that $\ln(1+t)=t+o(t)$ as $t\to 0$.
Hence the required limit is $e^{-1}$.
A: 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 $e^{-1}.$
A: Use the following famous Stirling formula: Given $x>0$
$$ \lim_{x\to +\infty} \frac{\Gamma(x+1)}{\left(\frac{x}{e}\right)^x \sqrt{2x} }=\sqrt{\pi}. $$
Where $\Gamma$ is the bGamma function of Euler and $n! =\Gamma(n+1)$ for $n\in \mathbb{N}$
