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

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

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.

• 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 1:46
• 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. – marty cohen May 10 '16 at 2:08
• @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...) – Chip May 10 '16 at 2:16
• 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 2:39
• 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. – mathguy May 10 '16 at 2:51

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.

• 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 8:49
• 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

Here I show that $$1.9 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

and for you question:

$$2.9