How do I calculate this limit: $\lim\limits_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}}$?

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.

Solution 1:

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!