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I want to build this infinite continued fraction

$$F_{n}(x)= \frac{1}{1-x\frac{(n+1)^2}{4(n+1)^2-1}F_{n+1}(x)} $$

which gives for $n=0$

$$F_{0}(x)=\dfrac{1}{1-\dfrac{(1/3)x}{1-\dfrac{(4/15)x}{1-\dfrac{(9/35)x}{1-\ddots}}}}$$ I took inspiration from this Post (@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?

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    $\begingroup$ In your definition you should change + to -, that is iF2[n_]:={-x(n-1)^2/(4(n-1)^2-1),1}; $\endgroup$ – yarchik Jul 6 at 14:18
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You could use Nest to compute the continued fraction

Nest [# /. F[n_] :> 1/(1 - x (n + 1)^2/(4 (n + 1)^2 - 1) F[n + 1]) &, F[0], 5] 
(*1/(1 - x/(3 (1 - (4 x)/(15 (1 - (9 x)/(35 (1 - (16x)/(63 (1 - 25/99 x F[5])))))))))*)
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  • $\begingroup$ @Ulrich_Neumann, how to get the value for n=200000? $\endgroup$ – Gallagher Aug 31 at 18:55
  • $\begingroup$ @Gallagher Just replace "5" by "200000" ...But I'm afraid, wether Mathematica ever finishs the analytical calculation $\endgroup$ – Ulrich Neumann Sep 1 at 9:41

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