Yesterday I posted this conjecture, but then deleted it thinking it was false. Turns out Python doesn't define $a^b$ as a^b, but rather as a**b.
Conjecture: Denote by $G$ Catalan's constant, then $$G=\cfrac{1}{1+\cfrac{1^4}{8+\cfrac{3^4}{16+\cfrac{5^4}{24+\cfrac{7^4}{32+\cfrac{9^4}{40+\ddots}}}}}}$$
Given the connection $G$ has with the number $8$ shown here, as well as this continued fraction reaching nearly the first five decimal places of $G$ after around $200$ iterations (vinculums), I am confident this is true.
However, I do not know how to code a continued fraction on Python or Pari/GP (a friend of mine gave it a go, but also to no avail) up to an iteration $n$ without having to write it out manually, which is really tedious. Here is some python code from a friend, coding this fraction up to $12$ iterations to be $\approx 0.9151$, reaching the first three decimal places of $G$.
The only 'local' behaviour that I can say about continued fractions is that most of them are convergent, and that they all converge via oscillation at each iteration. But, more importantly, I'd like to know that if this be true, can it be shown from here that $G$ is irrational (or even transcendental, if you are willing)? I am aware this is an unsolved problem, which was what inspired me to write $G$ in another closed form.
Any thoughts?
Thank you in advance.
Best Answer
The accepted answer is misleading. The continued fraction may well be found in that reference, but this is not a result from 2002, but rather a trivial consequence of Euler's continued fraction formula from 1748. You should take a look at the wikipedia page:
https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula
Euler's continued fraction formula $$a_0 + a_0 a_1 + \ldots + a_0 \cdots a_n = \frac{a_0}{\displaystyle{1 - \frac{a_1}{\displaystyle{1 + a_1 - \frac{a_2}{\ldots (1 + a_{n-1}) - \frac{a_n}{1 + a_n}}}}}}$$
Now exactly as in the worked example in the wikipedia page for $\tan^{-1}(x)$, you get the completely formal identity: $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}} {(2n+1)^2} = x + x \left(\frac{-x^2}{3^2}\right) + x \left(\frac{-x^2}{3^2}\right) \left(\frac{-3^2 x^2}{5^2}\right) + x \left(\frac{-x^2}{3^2}\right) \left(\frac{-3^2 x^2}{5^2}\right)\left(\frac{-5^2 x^2}{7^2}\right)+ \ldots$$ $$=\frac{x}{\displaystyle{1 + \frac{x^2}{\displaystyle{9 - x^2 + \frac{(9x)^2}{25 - 9 x^2 + \displaystyle{ \frac{(25 x)^2}{49 - 25 x^2 + \ldots }}}}}}}$$
The case $x=1$ is your example. You can plug in $x=i$ if you want to get a continued fraction for $\pi^2/8$.
There are literally thousands of completely trivial continued fractions on can create in this way; take any infinite sum and just formally write out the corresponding Euler continued fraction, clearing denominators in the obvious way. None of those should be considered anything more than a corollary of Euler's result (given the evaluation of the initial sum). Of course, in this case, the evaluation of the initial sum is that it is $G$ by definition.
(And no, this doesn't give anywhere near good enough convergents to say anything about the rationality or otherwise of $G$.)