Add, Multiply and Divide to infinity

First let us deal with the continued fraction,

let,

$\begin{aligned} x= \left(\cfrac{1}{2n-\cfrac{1}{4n-\cfrac{1}{4n-\cfrac{1}{4n-_\ddots}}}}\right)\end{aligned}$

we have ,

$\dfrac{1}{x}= \left(2n-\cfrac{1}{4n-\cfrac{1}{4n-\cfrac{1}{4n-_\ddots}}}\right)\Rightarrow (2n-\dfrac{1}{x})=\dfrac{1}{ (2n+\dfrac{1}{x})}\\ \Rightarrow (4n^2-1)=\dfrac{1}{x^2}\\ \Rightarrow x^2=\dfrac{1}{(4n^2-1)}$

The problem can be restated as,

$\dfrac{\displaystyle\prod_{n=1}^{\infty}\left(1+x^2\right)}{\displaystyle\sum_{n=1}^{\infty} \left(x\right)^2}= \dfrac{\displaystyle\prod_{n=1}^{\infty}\left(1+\dfrac{1}{(4n^2-1)}\right)}{\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{(4n^2-1)}\right)}$

let us deal with the denominator first,

$\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{(4n^2-1)}\right)=\dfrac{1}{2}\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{(2n-1)}-\dfrac{1}{(2n+1)}\right)=\dfrac{1}{2}$

The numerator is,

$\displaystyle\prod_{n=1}^{\infty}\left(1+\dfrac{1}{(4n^2-1)}\right)=\displaystyle\prod_{n=1}^{\infty}\left(\dfrac{4n^2}{(4n^2-1)}\right)$

Consider the infinite product expansion of sine,

we have, $\dfrac{sin(x)}{x}=\displaystyle\prod_{n=1}^{\infty} \left(1-\dfrac{x^2}{n^2 \pi^2}\right)$

let $x=\dfrac{\pi}{2}$ , we get

$\dfrac{2}{\pi}=\displaystyle\prod_{n=1}^{\infty} \left(1-\dfrac{1}{(2n)^2}\right)=\displaystyle\prod_{n=1}^{\infty} \left(\dfrac{4n^2-1}{4n^2}\right)$

or $\displaystyle\prod_{n=1}^{\infty}\left(\dfrac{4n^2}{(4n^2-1)}\right)=\dfrac{\pi}{2}$

now $\dfrac{\displaystyle Numerator}{\displaystyle Denominator}=\dfrac{\dfrac{\pi}{2}}{\dfrac{1}{2}}=\large \pi=\boxed{3.1415}$