This question is related to at least five others:
This makes a short answer possible (and desirable).
For a numerical calculation backward recursion is proposed (again):
$$
a_{n-1} = \sqrt{1/2^n+a_n} \qquad \mbox{with} \quad \lim_{n\to\infty} a_n = 0
$$
Here comes the Pascal program snippet that is supposed to do the job:
program apart;
procedure again(n : integer);
var
a,two : double;
k : integer;
begin
two := 1;
for k := n downto 2 do
two := two/2;
a := 0;
for k := n downto 2 do
begin
a := sqrt(two+a);
two := two*2;
end;
Writeln(a);
end;
begin
again(52);
end.
Note that an error analysis is not implemented in the program. This has not much sense
because the accuracy is determined by the smallest $1/2^n$ that can be represented
with some significance; that is for $n\approx 52$ in double precision Pascal.
The outcome is, of course, in concordance with the value already found by
Lucian:
1.28573676335699E+0000
Disclaimer. I certainly would have tried the closed form
- whatever that means in modern times - if I only could believe that such
a thing does indeed exist here.