A limit with sum $S_n=1+\frac{n-1}{n+2}+\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}+\cdots +\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}\cdots \frac{1}{2n}$ $$S_n=1+\frac{n-1}{n+2}+\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}+\cdots     +\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}\cdots \frac{1}{2n}$$ Then $S_n/\sqrt{n}$ tends to $\frac{\pi}{2}$.
How to show then? the general term in $S_n$ is $$\frac{n-1}{n+2}\frac{n-2}{n+3}\cdots \frac{n-k}{n+k+1}=\frac{C_{n+1}^{k+2}}{C_{n+k+1}^{k+2}}$$ where $C_n^k=\frac{n!}{k!(n-k)!}$. Then how to do?

The logarithm of a term is roughly
$$\int_{n-k-\frac12}^{n-\frac12}\log t dt - \int_{n+\frac32}^{n+k+\frac32}\log t dt$$
Find the leading order behaviour of that, then approximate the sum $S_n by an integral.
