The main reason of this answer is that it's impossible to squeeze the content
below into a comment. Disclaimer: it's is only a **partial answer** to the
question as formulated. And it's no way better than the (IMHO final) comment by
Achille Hui.
Let the function $A(n)$ be defined by:
$$
A(n) = \sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}
$$
Numerical computation of the sequence in, for example, Pascal is quite simple:

function A(N : integer) : double;
var
w : double;
k : integer;
begin
w := 0;
for k := N downto 1 do
w := sqrt(1/k + w);
A := w;
end;

But the strange thing about it is that it's sort of wrong headed iteration. Instead
of going from $A_1$ to $A_n$ it goes from $A_{n+1}$ *down to* $A_1$:
$$
A_{n+1} = 0 \quad ; \quad A_k = \sqrt{\frac{1}{k} + A_{k+1}} \quad ; \quad 1 \le k \le n
$$
It's called *Backward Recursion* according to the internet (I've never seen it
before). So the question is, remarkably, to find $A$ as:
$$
A = \lim_{n\to \infty} A_1(n) \qquad \mbox{instead of} \qquad A = \lim_{n\to \infty} A_n
$$
The numerical outcome is, of course, in agreement with Achille's, far less accurate
though (what can be expected from double precision Pascal).