Finding value of $ \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$ 
Finding value of  $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$

Try: $$\lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \frac{2k}{2k+1} = \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \prod^{n}_{k=1}\frac{2k}{2k+1}$$
$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4 \cdot 6\cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}\times \frac{(2\cdot 4 \cdot 6 \cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}$$
$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4\cdot 6\cdot 2n)^4}{(1\cdot 2\cdot 3\cdots 2n)^2} = \lim_{n\rightarrow \infty}2^{4n}\cdot \frac{(n!)^4}{(2n!)^2}$$
Did not find any clue how to solve from that point
Could some help me to solve it, Thanks 

With Stirling's approximation $n!\sim\sqrt{2\pi n}(n/e)^n$ we get:
\begin{align}
\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}
&=\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\\
&=\prod^{n}_{k=1}\frac{(2k)^2}{(2k-1)(2k)}\cdot\frac{(2k)^2}{(2k)(2k+1)}\\
&=4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k-1)(2k)}\cdot 4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k)(2k+1)}\\
&=2^{4n}\cdot\frac{(n!)^2}{(2n)!}\cdot\frac{(n!)^2}{(2n+1)!}\\
&=2^{4n}\cdot\frac{(n!)^4}{((2n)!)^2(2n+1)}\\
&\sim 2^{4n}\cdot\frac{(\sqrt{2\pi n}(n/e)^n)^4}{(\sqrt{2\pi 2n}(2n/e)^{2n})^2(2n+1)}\\
&\sim 2^{4n}\cdot\frac{4\pi^2n^2(n/e)^{4n}}{4\pi n(2n/e)^{4n}(2n)}\\
&\sim 2^{4n}\cdot\frac{\pi n^2}{2n^2 2^{4n}}\\
&\xrightarrow{n\to\infty}\frac\pi 2\\
\end{align}
