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$$\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$$

I have no idea about this.

The equation can be written in its recursive form as:

$$f(n) = g(1,n)$$

Where

$$g(x,n) = [x\impliedby n]\cdot (x+ g(x+1,n))^{\frac 1x}+[x=n]\cdot (n)^{\frac 1n}$$

Of course, [] is the indicator function representing of piece wise notation.

Mr Pie
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user338371
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    The computer gives the following: $$ 2: , 2.41421356237309 \
    3: , 2.85533004349830 \ 4: , 2.90798456765468 \ 5: , 2.91148304056081 \ 6: , 2.91163449677407 \ 7: , 2.91163911038987 \ 8: , 2.91163921441793 \ 9: , 2.91163921622082 \ 10: , 2.91163921624555 \ 11: , 2.91163921624582 \ 12: , 2.91163921624582 $$ ($n$ and $f(n)$)     – Chip May 10 '16 at 01:46
  • The RILYBOT Inverse Equation Solver at http://mrob.com/pub/ries/ries.php?target=2.91163921624582&rst= has a number of expressions close to this (within 1e-7), but none look promising. – marty cohen May 10 '16 at 02:08
  • @martycohen: indeed, I tried also the inverse calculator http://isc.carma.newcastle.edu.au/index and they couldn't identify the value (would be useful if somebody here can do a double check on the computer results, just to make sure...) – Chip May 10 '16 at 02:16
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    Do you have any reason to believe there is a simpler expression for this limit other than the one that you have given? – Eric Wofsey May 10 '16 at 02:39
  • The interesting problem is to show this sequence has a finite limit. It is strictly increasing, so the issue is to show it is bounded. –  May 10 '16 at 02:51
  • @mathguy it is not that hard to show it is bounded, it is clear the terms of the sequence will become bigger if we replace all $\sqrt[n]{\cdots}$ by $\sqrt{\cdots}$. By Herschfeld's convergence theorem, the bigger sequence converges. – achille hui May 10 '16 at 02:58
  • if it inspires anyone in this post, here is a graph of $f(n) = (n+(n+1+(\cdots))^\frac{1}{n+1})^\frac{1}{n}$. http://i.stack.imgur.com/2tn16.jpg The $f(1)$ is the required limit in the question of this post. – Chip May 10 '16 at 03:04
  • @achillehui - of course, what is "hard" is a matter of opinion. I have a PhD in math (and an IMO medal) and I have never heard of Herschfeld's theorem. Of course, replacing the nth roots with square roots is how you show boundedness. I didn't say "hard" - I said "interesting"! –  May 10 '16 at 03:09

2 Answers2

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Summarizing the above comments: It is clear that $$ x_n = 1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}} $$ is monotonically increasing, so that the convergence of this sequence is equivalent to its boundedness.

It has been demonstrated here that the sequence $$ y_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + …+\sqrt{n}}}} $$ is convergent. Since $x_n \le y_n^2$, this implies the convergence of $(x_n)$ as well.

The exact limit of $(y_n)$ is unknown, so that one can assume that the exact limit of $(x_n)$ is difficult to compute as well.

Martin R
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  • For all $n$, the number $x_n$ is algebraic of increasing degree but the limit is (informally speaking) of infinite "degree" and then it is a transcendent number. We could assure you that it not have a closed form. – Piquito Sep 25 '19 at 13:09
  • By the way, Mahler proved that the number $0.1234567891011…. $, whose decimal part follows the sequence of natural integers, is transcendent. – Piquito Sep 25 '19 at 16:14
  • @Piquito: I am not an expert on this topic, but does being transcendental prevent the number from having a “closed form”? – Martin R Sep 26 '19 at 08:49
  • No.For example $\cos (10)$ it is known to be trascendental But here is the form of each $x_n$ which leads me to such a statement (on conditional time though!). – Piquito Sep 27 '19 at 11:38
2

Here I show that $1.9<Lim_{n→∞} a_n<2$ if:

$a_n=\sqrt{2+\sqrt [3]{3+\sqrt[4]{4+ . . . +\sqrt[n]{n}}}}$

We use following inequalities:

$\sqrt[k]{k+1}>1$ ; $\sqrt[k]k<\sqrt[k]{k+2}$. . (for $k>2$)

The first one is clear and the second one can be proved by induction. The series of numbers of $a_n $is increasing; using inequalities we may write:

$a_n=\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n]{n}}}<\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n-1]{n-1+2}}}< . . .<\sqrt{2+\sqrt[3]{5}}$

Therefore aeries $a_n $ has a limit equal to $a_0$ such that we have:

$a_0<\sqrt{2+\sqrt[3]{5}}<2$

Using inequalities we have:

$a_n>\sqrt{2+\sqrt [3]{3+ . . . +\sqrt[n-1]{n}}}>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$

Therefore:

$a_0>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$

We can easily see that:

$\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}>1.9$

That finally gives:

$1,9<a_n<2$

and for you question:

$2.9<a_1+1<3$

sirous
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