How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$

Question

Question:

Show that $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$ exists, and find the best estimate limit $A$.

It is easy to show that

$$\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}\le\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$ and it is well known that this limit $$\sqrt{1+\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}}$$ exists.

So $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$

But can use some math methods to find an approximation to this $A$ by hand?

and I guess maybe this is true: $$1<A\le (\pi)^{\frac{1}{e}}?$$

By the way: we can prove $A$ is a transcendental number?

Thank you very much!

Answer 1

The nature and closed form expression of these two related constants, i.e., the Nested Radical Constant and Somos's Quadratic Recurrence Constant, are (also) unknown. This would suggest that the same holds true for this one as well, meaning that we are dealing with an open question.

As far as numeric approximations are concerned, $\displaystyle{A\simeq\frac{(\pi+1)\ln4}{1+\ln16}}$ comes close, within an error of less than $10^{-8}$.

Answer 2

Well, you can get very good approximations by computing:

$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$

---

$$b_1=\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}...}}}=\frac{1+\sqrt{3}}{2}$$

$$b_2=\sqrt{\frac{1}{3}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{3}...}}}=\frac{1+\sqrt{7/3}}{2}$$

$$b_3=\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}...}}}=\frac{1+\sqrt{2}}{2}$$

Then you use the facts that:

$$A^2-1<b_1$$

$$(A^2-1)^2-\frac{1}{2}<b_2$$

And so on, to calculate upper bounds for $A$ with very good accuracy:

$$A<\sqrt{1+b_1}=1.53819$$

$$A<\sqrt{1+\sqrt{\frac{1}{2}+b_2}}=1.52580$$

$$A<\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+b_3}}}=1.52300$$

$$A<\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+b_4}}}}=1.52224$$

And so on. This is already very close to the accurate expression. It's a rather easy calculation and can be done by hand if needed.

For the lower boundary you can use $b=1$ (it's rather obvious):

$$A>\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\sqrt{\frac{1}{4}+1}}}}=1.51844$$

Answer 3

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).