Does $A_n= \sqrt{1^2+\sqrt{2^2+\sqrt{...+\sqrt{n^2}}}}$ converge? Can we find the value of $\lim\limits_{n\to \infty} A_n$, does it converge?
$$A_n= \sqrt{1^2+\sqrt{2^2+\sqrt{3^2+\sqrt{4^2+\sqrt{...+\sqrt{n^2}}}}}} $$
I tried to calculate $A_1,A_2,\cdots A_{10}$, but my computer cannot do further terms.
 A: See here for references to a criterion.
$$\limsup_{n\to\infty} \frac{\log n^2}{2^n} = 2\limsup_{n\to\infty} \frac{\log n}{2^n} = 0$$
So the convergence is positive. The value is another story though.
A: Of course, we can do a numeric approximation. I used the following code (Mathematica):

N[Module[{n = 0}, 
         Nest[(n++; # /. x -> n^2 + Sqrt[x]) &, 
              Sqrt[x], 50]
  ] /. x -> 0,
  50]

which first recursively puts an $x$ inside the $51$st radical, then replaces it by zero, and approximates the result to $50$ digits.
This gives the number:
$$A_{50} \approx 1.9426554227639873282214132914126672376880736300071$$
The above code can easily be executed for deeper nested radicals to obtain more precision, but the first $50$ digits are correct.
The Inverse Symbolic Calculator does not have any suggestions for the limit -- not even in "advanced" mode. So it may just be your average irrational number outside of $\mathsf{Alg}(\Bbb Q[\pi,e])$, the numbers algebraic over $\Bbb Q[\pi,e]$.
