76
$\begingroup$

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!

$\endgroup$
  • 9
    $\begingroup$ Numerically, the number $\sim 1.5218903868642315049804187356992561937433633828577161062431190002880321340$ which is very different from $\pi^{1/e}$. $\endgroup$ – achille hui Nov 21 '13 at 16:24
  • 3
    $\begingroup$ Numerically, we should give up on trying to make any conclusions. $\endgroup$ – Ahaan S. Rungta Nov 21 '13 at 16:42
  • 4
    $\begingroup$ How are you so certain that $A$ is transcendental? $\endgroup$ – Steven Stadnicki Nov 21 '13 at 16:44
  • 8
    $\begingroup$ I don't get it. The OP has a proof that the limit exists, and achille hui has calculated the limit to $73$ decimal places. Is there anything more to do here? $\endgroup$ – Jim Belk Nov 22 '13 at 2:39
  • 6
    $\begingroup$ Yea, find closed form or prove it doesnt exist. How fast does this thing converge? Generally things which depend heavily on the first few terms dont have closed forms. $\endgroup$ – Abdulh Khazzak Gustav ElFakiri Nov 22 '13 at 9:33
16
$\begingroup$

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 $10^{-8}$.

$\endgroup$
  • $\begingroup$ You refer to nested square roots of $1$ or $x$. This isn't the case $\endgroup$ – Riccardo.Alestra Nov 21 '13 at 17:36
  • 6
    $\begingroup$ The two articles inform us of the following facts: $(1)$. Such expressions are not currently known to possess a closed form. $(2)$. Their nature is yet unknown. Yes, everybody suspects them of being transcendental: But maths is based on demonstration, not suspicion. $\endgroup$ – Lucian Nov 21 '13 at 17:59
  • $\begingroup$ I'm interested: how did you derive your approximation? (I ask because I wouldn't have a clue where to begin with deriving it...) $\endgroup$ – Alex Nelson Nov 23 '13 at 2:00
  • 6
    $\begingroup$ By using this site. $\endgroup$ – Lucian Nov 23 '13 at 2:06
5
$\begingroup$

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).

$\endgroup$
4
$\begingroup$

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):

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.