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

Question

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.

@AngelBlasco I dont know how to solve this because it is like unless I get the right idea i am not able to try much. — Sudhanshu, Jan 30 '16 at 12:38
Can you tell us where on the internet you found this question? — , Jan 30 '16 at 13:06
@Sudhanshu, was the question on quora to calculate the limit, or only to prove the limit exists? Can you link to where you found it? — Barry Cipra, Jan 30 '16 at 13:11
It was to calculate the limit but it had no answers as of when i viewed it. I will give the link soon. — Sudhanshu, Jan 30 '16 at 13:18
https://www.quora.com/How-do-I-calculate-lim-limits_-n-to-infty-1+-sqrt-2-2+-sqrt-3-3+-dotsb+-sqrt-n-n Heres the link — Sudhanshu, Jan 30 '16 at 15:19
And why those close votes ? I am just not able to even begin so I cannot show any working.. — Sudhanshu, Jan 30 '16 at 15:20
@Sudhanshu, thanks for the link. The poster there seems to acknowledge that there is likely no closed form for the limit. — Barry Cipra, Jan 30 '16 at 15:42
@BarryCipra I agree but since that was not a math forum and he just thought that there is no closed form . So there could be a chance that someone here could do it maybe ? — Sudhanshu, Jan 30 '16 at 16:26
Does it help to note: $\lim_{n\to\infty}\sqrt[n]{n}=1$? Perhaps it helps simplify or at least approximate? — Simply Beautiful Art, Jan 31 '16 at 13:44
How about binomial expansion? (I don't think it actually helps, but you can try) — Simply Beautiful Art, Jan 31 '16 at 15:40

score 6 · Answer 1 · answered Jan 30 '16 at 13:33

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!

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? — , 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 $$ \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{\ldots}}}} $$ which is the case $a=1$ of this previous Question. Wolfram Mathworld 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

Simply Beautiful Art · Answer 2 · 2016-01-31T14:19:48.617

0

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 $L$.

edited Jan 31 '16 at 14:19

answered Jan 31 '16 at 13:55

Simply Beautiful Art

74,685

Yes I tried similarly but most of the time I came back to the same thing. – Sudhanshu Jan 31 '16 at 14:00
hm.....well then. – Simply Beautiful Art Jan 31 '16 at 14:03
@Sudhanshu What about repeating the process more times, or maybe denesting methods? I'm not to familiar with those methods though. – Simply Beautiful Art Jan 31 '16 at 14:04
My trials seem so silly and I end up reversing my steps so it doesn't lead us anywhere. – Sudhanshu Jan 31 '16 at 14:10
@Sudhanshu That sucks. Round and round we go... – Simply Beautiful Art Jan 31 '16 at 14:16
What to do with your IDENTITY $L^2-2L+1=(L-1)^2$? It has the same value than, say, $L=L$ for instance. – Piquito Jan 31 '16 at 16:11
@Piquito I simply put it out there as an idea on how we could find $L$. I do indeed note that it is an identity, possibly useless. But the point of this was to try and spark ideas from readers like you. – Simply Beautiful Art Feb 01 '16 at 00:06
@SimpleArt: I confess that I had not read what you wrote after the identity. My apologies. – Piquito Feb 01 '16 at 14:06

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

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