I want to build this infinite continued fraction
$$F_{n}(x)= \cfrac{1}{1-x\cfrac{(n+1)^2}{4(n+1)^2-1}F_{n+1}(x)} $$
which gives for $n=0$
$$F_{0}(x)=\cfrac{1}{1-\cfrac{(1/3)x}{1-\cfrac{(4/15)x}{1-\cfrac{(9/35)x}{1-\ddots}}}}$$ I took inspiration from this post by @Michael E2, the problem is that when I transform it as a list representation
{b0,{a1, b1},{a2, b2},...}
Clear[F2,iF2];
iF2[0]=0;
iF2[1]={1,1};
iF2[2]={-x/3,1};
iF2[n_]:={-x(n+1)^2/(4(n+1)^2-1),1};
F2[n_]:=Table[iF2[k],{k,0,n}];
I can't find all the terms, so I find for 5 terms
Block[{n=5},F2[n]]
(*{0,{1,1},{-x/3,1},{-16x/63,1},{-25x/99,1},{-36x/143,1}}*)
it lacks after {$-x/3,1$} the terms {$-4x/15,1$} and {$-9x/35,1$}
What is wrong please?
iF2[n_]:={-x(n-1)^2/(4(n-1)^2-1),1};
$\endgroup$ – yarchik Jul 6 '19 at 14:18