$\sqrt {1+\sqrt {2+\ldots +\sqrt {n}}}< 2 , \forall n\ge 1$ $\sqrt {1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}}<\sqrt {2+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}} , \forall n\ge 1$ $\sqrt {2+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}}}<\sqrt {2+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}} , \forall n\ge 1$ Now we have $\sqrt {1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}}<\sqrt {2+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}}} , \forall n\ge 1$ And that is true because $1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}<1+1+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}} , \forall n\ge 1$ Is obviously that $\sqrt {2+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}}}<2$ And I prove that $\sqrt {1+\sqrt {2+\ldots +\sqrt {n}}}<\sqrt {2+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}}}< 2 , \forall n\ge 1$

enescu wrote: Prove that $\sqrt {1+\sqrt {2+\ldots +\sqrt {n}}}<2$, $\forall n\ge 1$. I think this problem is really well-known. I think I also have seen it on some "Tournament Of Town" paper. Probably our goldie is really loved by the mathematical community or it is to easy to rediscover it.

Lucmar wrote: Now we have $\sqrt {1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}}<\sqrt {2+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}}} , \forall n\ge 1$ And that is true because $1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}<1+1+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}} , \forall n\ge 1$ I am not sure I understand this argument. By the way, Orlando, I think you are refering to TOT problem 4 Spring 1987. But this is $\sqrt {2\sqrt {3\sqrt {4\ldots \sqrt {n}}}}<3.$

It all follows from $\sqrt{k+\sqrt{k+1+\sqrt{\ldots+\sqrt n}}}<k+1$ for $k\ge 1$. It can be proven by "induction", starting from $k=n$ and going down to $k=1$.

Yes, of course. The same approach works for the TOT problem.

enescu wrote: Lucmar wrote: Now we have $\sqrt {1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}}<\sqrt {2+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}}} , \forall n\ge 1$ And that is true because $1+\sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}<1+1+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}} , \forall n\ge 1$ I am not sure I understand this argument. I rise the equation to the 2nd power. I wonder if it is another way to do the problems other than "induction".

Lucmar wrote: I rise the equation to the 2nd power. That I understood. I still don't see why the inequality \[ \sqrt {2+\sqrt {3+\ldots +\sqrt {n}}}<1+\sqrt {2+\sqrt {2+\ldots +\sqrt {2}}} , \forall n\ge 1\] is that obvious.

, i thinks this pro very easy and we have pro: PRO : assume q possitive real number,define: $f_q({n})=(1+(2+(........+n^q)^q).....)^q)^q$,show that if $q<1$ then there exist $\lim{f_q({n})}$ i think lucmar have some mistake Sol:$\sqrt{n}<n$ so $\sqrt{n-1+\sqrt{n}}<\sqrt{2n-1}<n-1$ for $n-1>4$ and we obtain $\sqrt {1+\sqrt {2+\sqrt {n}}}<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2\cdot{4}-1}}}}<2$

enescu wrote: Prove that $$\sqrt {1+\sqrt {2+\ldots +\sqrt {n}}}<2$$$\forall n\ge 1$ For any natural $n$ prove the inequality $$\sqrt{2\sqrt{2}{\sqrt{3}\sqrt{4 ...\sqrt{n-1\sqrt{n}}}}} <3$$Tournament of Towns, Spring 1987