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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!

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    $\begingroup$ Numerically, the number $\sim 1.5218903868642315049804187356992561937433633828577161062431190002880321340$ which is very different from $\pi^{1/e}$. $\endgroup$ Nov 21, 2013 at 16:24
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    $\begingroup$ Numerically, we should give up on trying to make any conclusions. $\endgroup$ Nov 21, 2013 at 16:42
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    $\begingroup$ How are you so certain that $A$ is transcendental? $\endgroup$ Nov 21, 2013 at 16:44
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    $\begingroup$ 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, 2013 at 2:39
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    $\begingroup$ 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. $\endgroup$ Nov 22, 2013 at 9:33

3 Answers 3

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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 $10^{-8}$.

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  • $\begingroup$ You refer to nested square roots of $1$ or $x$. This isn't the case $\endgroup$ Nov 21, 2013 at 17:36
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    $\begingroup$ 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, 2013 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...) $\endgroup$ Nov 23, 2013 at 2:00
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    $\begingroup$ By using this site. $\endgroup$
    – Lucian
    Nov 23, 2013 at 2:06
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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).

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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):

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

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