Let $n \geq 2$. I need to show that $\sqrt{2\sqrt{3\sqrt{4...\sqrt{n}}}} < 3$ for all $n \in \mathbb{N}$.
I tried using logarithms and got arithmetico-geometric like series, except for the $\log$ operator. Can't think of a useful inductive argument here.
Asked
Active
Viewed 107 times
2
martianwars
- 357