Limit

[myra_panama]

$\[ for\;n\in{N_{+}},{S_{n}}= 1+\frac{{n-1}}{{n+2}}+\frac{{n-1}}{{n+2}}\cdot\frac{{n-2}}{{n+3}}+\ldots+\frac{{n-1}}{{n+2}}\cdot\frac{{n-2}}{{n+3}}\cdot\ldots\cdot\frac{1}{{2n}},prove\;\;\mathop{\lim }\limits_{n\to\infty }\frac{{{S_{n}}}}{{\sqrt n }}=\frac{\pi }2 \]$

[myra_panama]

Is there any idea?

[Vince Diesel]

Ага. Умножить $S_n$ на $C_{2n}^{n+1}$. Однако зачем в этот раздел? Хотите, чтобы за вас решили? Только ответ не совсем такой будет.

[Руст]

It is not hard, but long to solution. May be exist short way by use trigonometric methods.
Directly way^
$S_n=\sum_{k=0}^{n-1}r_k, r_0=1,r_k=\prod_{i=1}^k a_i, a_i=\frac{n-i}{n+i+1}$.
$\ln(r_k)=\sum_{i=1}^k \ln(1-\frac{i}{n})-\ln(1+\frac{i+1}{n})$.
If $k>n^{2/3}$, then [math][math][math]$$\sum_{k>n^{2/3}}r_k<nexp(-n^{1/3})\to 0$.
Therefore we can consider only members with $k\le n^{2/3}$. In this case

$\ln(r_k)=\frac{(k+1)^2-1}{n}+\frac{(k+1)^2-1}{2n^2}+O(\frac{k^4}{n^3}).$
Or $r_k=exp(-(k+1)^2/n)(1+O(n^{-2/3})$.
Therefore

$$S_n=\sum_{k=1}^{n^{2/3}}exp(-\frac{k^2}{n}) +O(1).$$

Let $x_k=\frac{k}{\sqrt n}$, then $S_n=\sqrt n \sum_k exp(-x_k^2)dx_k +O(1)=\sqrt n \int_0^{\infty}exp(-x^2)dx +O(1).$
It gives result.