Sequence problem regarding convergence from an online contest Let $(x_n)_{n\in \mathbb{N}}$ be a sequence defined by $x_0=1$ and $x_n=x_{n-1}\cdot \big(1-\frac{1}{4n^2}\big)$, $\forall n\geq 1$.
Prove that:
a) $(x_n)_{n\in \mathbb{N}}$ is convergent
b) if $l=\lim_{n\to \infty} x_n$, compute $\lim_{n \to \infty} (\frac{x_n}{l})^n$.
What I did was substitute $n-1,n-2,\ldots,1$ in the recurrence relation and I got that $x_n=\prod_{k=1}^{n} \big(1-\frac{1}{4k^2}\big)$. However, here I got stuck because I don't know how to find this limit. 
 A: Let $w_n={x_n}^{-1}.$
(a) $(w_n)$ is the Wallis sequence which converges to $\pi/2$. The sequence $(x_n)$ therefore converges to $2/\pi$.
(b) It is well-known that $(w_n)$ converges like $\frac {1}{n}$ but to answer part (b) we need a more precise result. One such result (Hirschhorn) is as follows:-
$$1-{\frac{1}{4n+\frac{7}{3}}}<\frac{2w_n}{\pi}<1-{\frac{1}{4n+\frac{8}{3}}}.$$
Then $(\frac{2w_n}{\pi})^n$ and  $(1-\frac{1}{4n})^n$ converge to the same limit i.e. $e^{-\frac{1}{4}}$.
The solution to part(b) is therefore $e^{\frac{1}{4}}$.
Reference
Sharp inequalities and asymptotic expansion associated with the Wallis sequence, Deng et al, Journal of inequalities and Applications (2015) 2015:186.
A: Partial Answer
We have $$x_n=\prod_{k=1}^{n} \big(1-\frac{1}{4k^2}\big){=\prod_{k=1}^n{(2k-1)(2k+1)\over (2k)^2}\\=\prod_{k=1}^n{(2k-1)\cdot 2k\cdot 2k\cdot(2k+1)\over (2k)^4}\\={1\over 16^n}\cdot {1\over(n!)^4}\prod_{k=1}^n(2k-1)\cdot 2k\cdot 2k\cdot(2k+1)\\={(2n)!\cdot (2n+1)!\over 16^n\cdot (n!)^4}}$$therefor by using the Stirling's approximation for factorial we obtain $$x_n{\sim{1\over 16^n}\cdot {\sqrt{4\pi n}({2n\over e})^{2n}\sqrt{4\pi n+2\pi}({2n+1\over e})^{2n+1}\over \sqrt{2\pi n}^4({n\over e})^{4n}}}$$when $n\to \infty$, we have $${\sqrt{4\pi n}\sqrt{4\pi n+2\pi }\over \sqrt{2\pi n}^4}\sim {1\over \pi n}$$and $${{1\over 16^n}{({2n\over e})^{2n}({2n+1\over e})^{2n+1}\over ({n\over e})^{4n}}={2n\over e}\cdot ({1+{1\over 2n}})^{2n+1}}\sim 2n$$by multiplying the former constituent terms of $x_n$ we have 

$$\lim_{n\to \infty}x_n={2\over \pi}$$

A: a) $\frac{x_{n+1}}{x_n}<1$ as $n$ grows. Also, $x_n>0$. So $x_n$ is a strictly decreasing bounded sequence and as a result it's convergent to $l= \inf\{x_n:n \in N\}$.  
