6

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.

Watson
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Steven Sun
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2 Answers2

3

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.

Community
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AlexR
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  • Can we find the vaule? – Steven Sun Sep 19 '13 at 11:11
  • The value can be found by bijection. Take $x_1$ as a suitable approximation of the limit and define: $$x_n = x_{n-1}^2 - (n-1)^2.$$ If, soon or later, $x_n\geq n+2$, then $x_1$ is an approximation from above of the limit, because we have $x_N\geq N+2$ for any $N\geq n$. Otherwise, if $x_m\leq m+1$ happens, $x_1$ is an approximation from below, because we have $x_{m+1}<0$. – Jack D'Aurizio Sep 19 '13 at 17:31
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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 $\Bbb Q[\pi,e]$.

Lord_Farin
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