16
$\begingroup$

Is there a closed form of this:

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

$\endgroup$
9
  • $\begingroup$ It is interesting . and can't see it.hope someone take some usefull [Continued fraction]:mathworld.wolfram.com/ContinuedFraction $\endgroup$
    – math110
    Commented Dec 19, 2014 at 11:36
  • $\begingroup$ not too far from $459/314$ $\endgroup$
    – Henry
    Commented Dec 19, 2014 at 11:47
  • 1
    $\begingroup$ $1.461783355000579602560079367$ is the value of the number. $\endgroup$
    – Peter
    Commented Dec 19, 2014 at 12:17
  • 3
    $\begingroup$ $\large \frac{55099}{37693}$ is an approximation with an error of abot $10^{-12}$ $\endgroup$
    – Peter
    Commented Dec 19, 2014 at 12:37
  • 1
    $\begingroup$ @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 ...) $\endgroup$
    – Zubin Mukerjee
    Commented Dec 19, 2014 at 17:00

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .