An Inequality with Nested Radicals

Prove that, for all integer $n\ge 2,$

$\displaystyle\sqrt{2\sqrt{3\sqrt{4\ldots\sqrt{n}}}}\lt 3.$

The solution by Dorin Marghidanu below serves a perfect illustration to G. Polya's maxim that The more ambitious plan may have more chances of success.



$\displaystyle R_{k,n}=\sqrt{k\sqrt{(k+1)\sqrt{(k+2)\ldots\sqrt{n}}}}.$

We'll prove that $R_{k,n}\lt k+1.$ Starting with $k=n-1$ we then descend to $k=2:$

$\begin{align} R_{n-1,n}&=\sqrt{(n-1)\sqrt{n}}\lt\sqrt{(n-1)n}\lt n,\\ R_{n-2,n}&=\sqrt{(n-2)R_{n-1,n}}\lt\sqrt{(n-2)n}\lt \sqrt{(n-1)^{2}-1}\lt n-1,\\ R_{n-3,n}&=\sqrt{(n-3)R_{n-2,n}}\lt\sqrt{(n-3)(n-1)}\lt \sqrt{(n-2)^{2}-1}\lt n-2,\\ &\cdots\\ R_{k,n}&=\sqrt{k\cdot R_{k+1,n}}\lt\sqrt{k(k+2)}\lt \sqrt{(k+1)^{2}-1}\lt k+1,\\ &\cdots\\ R_{2,n}&=\sqrt{2\cdot R_{3,n}}\lt\sqrt{2\cdot 4}\lt \sqrt{3^{2}-1}\lt 3.\\ \end{align}$


For each fixed $k\ge 2,$ $R_{k,n}$ is monotone increasing. Since $R_{k,n}$ is bounded, independent of $n,$ the sequences $\{a_{n}\},$ $a_{n}=R_{k,n}$ each have a limit, say, $L_{k}.$ What are these?

|Contact| |Front page| |Contents| |Algebra| |Generalization|

Copyright © 1996-2018 Alexander Bogomolny