6
$\begingroup$

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

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.

$\endgroup$
  • 3
    $\begingroup$ 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 $f(n)$) $\endgroup$ – Chip May 10 '16 at 1:46
  • $\begingroup$ The RILYBOT Inverse Equation Solver at 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. $\endgroup$ – marty cohen May 10 '16 at 2:08
  • $\begingroup$ @martycohen: indeed, I tried also the inverse calculator 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...) $\endgroup$ – Chip May 10 '16 at 2:16
  • 4
    $\begingroup$ Do you have any reason to believe there is a simpler expression for this limit other than the one that you have given? $\endgroup$ – Eric Wofsey May 10 '16 at 2:39
  • $\begingroup$ 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. $\endgroup$ – mathguy May 10 '16 at 2:51
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Piquito Sep 25 '19 at 13:09
  • $\begingroup$ By the way, Mahler proved that the number $0.1234567891011…. $, whose decimal part follows the sequence of natural integers, is transcendent. $\endgroup$ – Piquito Sep 25 '19 at 16:14
  • $\begingroup$ @Piquito: I am not an expert on this topic, but does being transcendental prevent the number from having a “closed form”? $\endgroup$ – Martin R Sep 26 '19 at 8:49
  • $\begingroup$ 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!). $\endgroup$ – Piquito Sep 27 '19 at 11:38
2
$\begingroup$

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:

$2.9<a_1+1<3$

$\endgroup$

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.