How to prove this determinant is $\pi$?

prove or disprove $$\pi=\begin{vmatrix} 3&1&0&0&0&\cdots\\ -1&6&1&0&0&\cdots\\ 0&-1&\dfrac{6}{3^2}&1&0&\cdots\\ 0&0&-1&\dfrac{3^2\cdot 6}{5^2}&1&\cdots\\ 0&0&0&-1&\dfrac{5^2\cdot 6}{3^2\cdot 7^2}&\cdots\\ 0&0&0&0&-1&\dfrac{3^2\cdot 7^2\cdot 6}{5^2\cdot 9^2}&\cdots\\ 0&0&0&0&0&-1&\dfrac{5^2\cdot 9^2\cdot 6}{3^2\cdot 7^2\cdot 11^2}&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\ddots\\ \end{vmatrix}$$

I found maybe this is true.and is very interesting,(It seems Euler proved it?),because this follows from the Euler result:

$$\pi=3+\dfrac{1^2}{6+\dfrac{3^2}{6+\dfrac{5^2}{6+\dfrac{7^2}{6+\cdots}}}}$$

and can we solve it? Thank you

Solution 1:

A proof can be found on pages 11-13 of An Elegant Continued Fraction for $\pi$. Try deriving from the Leibnitz formula $1 - \frac{1}{3} + \frac{1}{5} - ... = \frac{\pi}{4}$