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Is there a closed form of this:

$$ 1!+\dfrac{1}{2!+\dfrac{1}{3!+\dfrac{1}{4!+\ldots}}} $$

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  • $\begingroup$ It is interesting . and can't see it.hope someone take some usefull [Continued fraction]:mathworld.wolfram.com/ContinuedFraction $\endgroup$
    – math110
    Dec 19, 2014 at 11:36
  • $\begingroup$ not too far from $459/314$ $\endgroup$
    – Henry
    Dec 19, 2014 at 11:47
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    $\begingroup$ $1.461783355000579602560079367$ is the value of the number. $\endgroup$
    – Peter
    Dec 19, 2014 at 12:17
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    $\begingroup$ $\large \frac{55099}{37693}$ is an approximation with an error of abot $10^{-12}$ $\endgroup$
    – Peter
    Dec 19, 2014 at 12:37
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    $\begingroup$ @ajotatxe Careful with that bet or you might end up like Robespierre! (I guess "almost all" real numbers are transcendental, so you might be safe ...) $\endgroup$ Dec 19, 2014 at 17:00

1 Answer 1

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My 2 cents worth. Here is a Pascal program snippet that does Peter's job; actually half of it due to Delphi's double precision limitations. Backward recursion is the clue (again).

program Peter;
procedure fraction; var a : double; f,k : integer; begin f := 1*2*3*4*5*6*7*8*9*10; k := 10; a := f; while k > 1 do begin f := f div k; a := 1/a+f; k := k-1; end; Writeln(a); end;
begin fraction; end.
Output:
 1.46178335500058E+0000
Closed form? I don't think so.

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