1

I'm trying to prove that

$$ \sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}} < 3 $$

for any $n$ and have decided to use strong induction and instead just show that

$$ \sqrt{k\sqrt{(k + 1)\cdots\sqrt{n}}} < k + 1 $$

for any choice of $k$. Does anyone have an idea on where to proceed from here?

Lord_Farin
  • 17,924
  • 9
  • 52
  • 132
Michael Davidson
  • 29

3 Answers3

2

You might consider this.
$$ \sqrt{2\sqrt{3\sqrt{\cdots\sqrt n}}} = 2^{1/2}\cdot 3^{1/4} \cdots n^{1/2^n} =\exp\left\{\sum_{k=2}^n {\log(k)\over 2^{k-1}} \right\}$$ Can you bound the sum inside. It is convergent?

ncmathsadist
  • 50,127
1

Fix $n$. We prove by (downward) induction on $k$ ($k \le n$) that:

$$\sqrt{k\sqrt{(k+1)\cdots\sqrt n}} < k+1$$

For $k=n$ it is evident. Given $\sqrt{(k+1)\cdots \sqrt n}<k+2$, note that:

$$\frac12(k +(k+2)) = k+1 \ge \sqrt{k(k+2)}$$

by the AM-GM inequality. Use the induction hypothesis to conclude.

Lord_Farin
  • 17,924
  • 9
  • 52
  • 132
1

For every $1\leqslant k\leqslant n$, let $x_{k,n}=\sqrt{k\sqrt{(k + 1)\cdots\sqrt{n}}}$. Then $x_{k,n}^2=kx_{k+1,n}$ hence if $x_{k+1,n}\lt k+2$ then $x_{k,n}^2\lt k(k+2)=(k+1)^2-1\lt(k+1)^2$ hence $x_{k,n}\lt k+1$. The downward recursion on $k$ such that $1\leqslant k\leqslant n$, for any fixed value of $n\geqslant1$, is complete once one notes that $x_{n,n}=\sqrt{n}\lt n+1$ because $(n+1)^2-n=n^2+n+1\geqslant3\gt0$.

Did
  • 284,245