I just recently came across: $$\lim_{N\to \infty}\bigg[2\prod_{k=1}^{N}\biggl(\frac{(2k)^2}{(2k)^2-1}\biggr)\biggr]$$ which appears to be approaching $\pi$...
Can anyone explain to me why this is, or give me an article that explains why this is? My first thoughts were to rewrite this with factorials or something but I'm not sure that would make it obvious?
There are some "elementary" ways to show this, the one that is maybe simpler to type here is the following one. Consider the integrals $$ J_n:=\int_0^\pi \sin^n x\; dx\ ,\qquad n\in\Bbb N\ . $$ It turns out that $J_n$ satisfies a two step recurrence relation, which is relating $J_n$ with $J_{n-2}$. Explicitly, separating the cases of an even $n$, and of an odd $n$, we get $$ \begin{aligned} J_{2n} &= \underbrace{J_0}_{\pi}\cdot\frac 12\cdot \frac34\cdot\dots\cdot\frac{2n-1}{2n}\ , \\ J_{2n+1} &= \underbrace{J_1}_{2}\cdot\frac 23\cdot \frac45\cdot\dots\cdot\frac{2n}{2n+1}\ . \end{aligned} $$ Now odd index integral $J_{2n}$ "sits between" the even index integrals $J_{2n+1}$ and $J_{2n-1}$. And similarly, the even index integral $J_{2n+1}$ sits... The sandwich criterion tells us that the quotient $J_{2n}/J_{2n+1}$ goes to one for $n\to\infty$. Using the above formulas, (and neglecting some "easy part" which goes to one) we obtain: $$ 1 =\lim_{n\to\infty} \frac 2\pi \ \cdot\ \frac 21\cdot \frac 23 \ \cdot\ \frac 43\cdot \frac 45 \qquad \dots\qquad \ \cdot\ \frac {2n}{2n-1}\cdot \frac {2n}{2n+1}\ . $$ This is equivalent to the product in the quesition. (Details can be found on the net by searching for an elementary proof of Wallis' formula...)
(This is the way to remember why $\pi$ and $2$ occur in the final formula.)
Beside the Wallis product, what you can do is $$P_n=2\prod_{k=1}^{n}\frac{4k ^2}{4k^2-1}=2 \frac{\prod_{k=1}^n (4k^2) } {\prod_{k=1}^n (4k^2-1)}$$ $$\prod_{k=1}^n 4k^2=4^n \Gamma (n+1)^2$$ $$\prod_{k=1}^n (4k^2-1)=\frac{2^{2 n+1} \Gamma \left(n+\frac{1}{2}\right) \Gamma \left(n+\frac{3}{2}\right)}{\pi }$$ Combining all the above $$P_n=\pi \,\frac{ \Gamma (n+1)^2}{\Gamma \left(n+\frac{1}{2}\right) \Gamma \left(n+\frac{3}{2}\right)}$$ Now, take the logarithms and use Stirling approximation to obtain $$\log \left(\frac{\Gamma (n+1)^2}{\Gamma \left(n+\frac{1}{2}\right) \Gamma \left(n+\frac{3}{2}\right)}\right)=-\frac{1}{4 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, using $x=e^{\log(x)}$ $$P_n=\pi \left(1-\frac{1}{4 n}+\frac{5}{32 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ from which $$\frac{P_{n+1}}{P_n}=1+\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right) $$
Now, if you want to know $n$ such that $$\pi - P_n \leq \epsilon$$ you have $$n \geq \left\lceil \frac{\pi }{4 \epsilon }\right\rceil$$
