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I have seen this question on the internet and was interested to know the answer.

Here it is :

Calculate $\lim\limits_{n\to\infty}(1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}})$?

Edit : I really tried doing it but wasn't able to get somewhere.

I know how to do questions like $ y = (1+\sqrt{1+\sqrt{1+\dotsb+\sqrt 1}}) $ and then we write $ (y-1)^2 = y $ and solve.

But for this I have no method. So I would like even a sort of a hint to try to get me started, no need for answer.

Martin Sleziak
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Sudhanshu
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2 Answers2

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I know this is not what you asked, but since it's one step closer to the answer, I'll post it anyway. I'll prove that the limit exists.

First, given that $a<2$, we know that $2^n>n+a$ for all $n\geq 2$. We'll prove this by induction. First of all, it works for $n=2$, since $2^2>2+a$ is true, because we assumed $a<2$. Now given that $2^n>n+a$ for some $n$, we know that $$2^{n+1}=2\cdot 2^n>2n+2a>n+n+a>(n+1)+a$$ so it must be true for all $n$. From that, we deduce $\sqrt[n]{n+a}<2$. Since also $\sqrt[n]{n}<2$, we can use this fact to find an upper bound, since:\begin{align} a_n=\sqrt[n]{n}&<2\\ a_{n-1}=\sqrt[n-1]{n-1+a_n}&<2\\ a_{n-2}=\sqrt[n-2]{n-2+a_{n-1}}&<2\\ &\vdots\\ a_2=\sqrt[2]{2+a_3}&<2 \end{align} So the sequence will always be less than $1+2=3$. It is not so hard to see that the sequence is always positive, since roots of positive numbers are positive. It is also not too hard to see that is is monotone, since for en extra term you add a term to the "deepest lying" root, and so the entire thing will get more, too. I realize that the argument for the fact that it is an increasing sequence isn't very rigorous, but it visualizes it, and could also be made quite rigorous indeed.

Also, I computed the value for $n=1000$ with Wolfram Mathematica 10.0 and the value it gave was $2.9116392162458242839\cdots$

I hope this helped!

vrugtehagel
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    The ISC (inverse symbolic calculator) says "Wow, really found nothing." This suggests there is no compact expression for the limit. – hardmath Jan 31 '16 at 02:14
  • @hardmath, oh well, then I guess we should try to prove it is transcendental? – vrugtehagel Jan 31 '16 at 07:14
  • That seems both very plausible, but also technically difficult. – hardmath Jan 31 '16 at 12:14
  • Another upper bound (and hence proof of convergence) can be given by comparison with [the square of the Kasner number](https://oeis.org/A072449) $$ \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{\ldots}}}} $$ which is the case $a=1$ of [this previous Question](http://math.stackexchange.com/questions/48503/nested-radicals-sqrta-sqrt2a-sqrt3a-ldots). [Wolfram Mathworld](http://mathworld.wolfram.com/NestedRadicalConstant.html) calls it simply the Nested Radical Constant. There is also no known closed form for this, but its convergence follows from Vijayaraghavan's theorem. – hardmath Jan 31 '16 at 17:01
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Well, this is my best shot:

$$L=\lim_{n\to\infty}(1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}})$$

$$L^2=\lim_{n\to\infty}(1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}})^2$$

$$=\lim_{n\to\infty}1+2\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}}+2+\sqrt[3]{3+\dotsb+\sqrt[n]n}$$

$$L^2=\lim_{n\to\infty}1+2L+\sqrt[3]{3+\dotsb+\sqrt[n]n}$$

$$L^2-2L-1=\sqrt[3]{3+\dotsb+\sqrt[n]n}$$

$$L^2-2L+1=2+\sqrt[3]{3+\dotsb+\sqrt[n]n}$$

$$\sqrt{L^2-2L+1}=\sqrt{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}}$$

$$\sqrt{L^2-2L+1}+1=1+\sqrt{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}}$$

$$\sqrt{L^2-2L+1}+1=L$$

$$L^2-2L+1=(L-1)^2$$

Well, I tried. Turns out this is a true statement for all $L$, so I haven't narrowed it down. But perhaps my method raises ideas....

This is equivalent to trying to solve:

$$\lim_{n\to\infty}((\dots((L-1)^2-2)^3-3\dots)^n-n=0$$

And through a visual graph, we see the limit converges rapidly and only for some $L$.

Simply Beautiful Art
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