# Why $\forall n\gt2\in\Bbb N:\lfloor\sqrt{n\cdot\sqrt{n\cdot\lfloor\sqrt{…\lfloor\sqrt{n\cdot\lfloor\sqrt{n}\rfloor}}\rfloor\rfloor}}\rfloor=n-2$?

I have found this just by chance, but can not understand why the limit turns out to be $$n-2, \forall n\gt 2$$.

Def:

$$a_0 = \lfloor \sqrt{n} \rfloor$$

$$a_1 = \lfloor \sqrt{n \cdot a_0} \rfloor$$

...

$$a_k = \lfloor \sqrt{n \cdot a_{k-1}} \rfloor$$

And apparently:

$$\forall n\gt 2:\ lim_{k\to \infty} a_k = \lfloor \sqrt{n \cdot \sqrt{ n \cdot \lfloor \sqrt{...\lfloor \sqrt{n\cdot \lfloor \sqrt{n} \rfloor }} \rfloor \rfloor}} \rfloor = n-2$$

I understand that $$(n-2) \cdot n = n^2-2n \lt (n-1)^2 = n^2+1-2n$$ and for that reason if one of the elements of the sequence is $$n-2$$ then can not grow up more because $$\lfloor \sqrt {n \cdot (n-2)} \rfloor = n-2$$. But what I do not understand is why the sequence exactly arrives to $$n-2$$. Why not $$n-1$$ or $$n-3$$ for instance?

I would like to ask the following questions:

1. Is the observation correct? Are there counterexamples?

2. Why does exactly arrive to $$n-2$$ instead of any other value like, for instance, $$n-3$$? Probably the reason is quite trivial but I can not see the property behind the behavior. Thank you!

All $n\gt 2$ converge to $n-2$ as you surmise.
The easiest way to see it is to look what happens if $a_n=n-3$. Then $a_{n+1}=\lfloor \sqrt {n(n-3)} \rfloor = \lfloor \sqrt {n^2-3n} \rfloor$. If $n \ge 4, n^2-3n \ge n^2-4n+4 = (n-2)^2$ so you will climb from $n-3$ to $n-2$ As you have pointed out, once you get to $n-2$ you have stability because $n^2-2n \lt n^2-2n+1=(n-1)^2$ This argument does not work for $n=3,$ but $\lfloor \sqrt 3 \rfloor =1$ and we start out at $n-2$
• thank you for the explanation, I like the word you have wisely used: "stability". Just one point: I am specifying in the first paragraph $\forall n \gt 2$ so when I ask for a counterexample I meant for $n \gt 2$. – iadvd Sep 20 '16 at 3:21
• I missed that about $n \gt 2$. I will fix it. – Ross Millikan Sep 20 '16 at 3:22