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Prove that $\sqrt{2\sqrt{3\sqrt{...\sqrt{(n - 1)\sqrt{n}}}}} < 3$.

I have tried induction on $n$, but with no effect.

user128409235
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  • If you just try induction on $n$ directly, it won't work, because there is no way you can get from just $\sqrt{2\sqrt{3\sqrt{...\sqrt{(k - 1)\sqrt{k}}}}} < 3$ to $\sqrt{2\sqrt{3\sqrt{...\sqrt{k\sqrt{k+1}}}}} < 3$. There just isn't any room to work with. The standard solution is to actually prove something stronger in order for the induction step to work, something like $\sqrt{2\sqrt{3\sqrt{...\sqrt{(n - 1)\sqrt{n}}}}} < 3 - \frac1n$, or ${}<3-2^{-n}$, or something. I have no idea what would actually work here, though, so if you want to use induction, you will have to be clever. – Arthur Feb 02 '18 at 08:43
  • Well according to Ramanujan, $$1\cdot\sqrt{1 + 2\cdot\sqrt{1 + 3\cdot\sqrt{1 + \Lambda}}} = 3$$ such that $\Lambda$ represents that the nested radical continues forever. Perhaps this may be helpful? – Mr Pie Feb 02 '18 at 09:00
  • Proof: We begin by asserting the following: $$\begin{align} 3 &= \sqrt 9 = \sqrt{1 + 8} = \sqrt{1 + 2\cdot 4} = \sqrt{1 + 2\sqrt{16}} \ &= \sqrt{1 + 2\sqrt{1 + 15}} = \sqrt{1 + 2\sqrt{1 + 3\cdot 5}} \ &= \sqrt{1 + 2\sqrt{1 + 3\sqrt{25}}} = \cdots = \text{as stated.}\tag*{$\Box$}\end{align}$$ – Mr Pie Feb 02 '18 at 09:26
  • see answers of this. – achille hui Feb 02 '18 at 09:27

1 Answers1

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You can do a downward induction: $$r(k,n)=\sqrt{k\sqrt{(k+1)\sqrt{...\sqrt{(n - 1)\sqrt{n}}}}} <k+1$$ for $k=1,\ldots,n$.
Proof: It's true for $k=n$, since we always have $\sqrt{n}<n+1$. If it's true for some $k$, it's also true for $k-1>0$: $$r(k-1,n)=\sqrt{(k-1)\,r(k,n)}<\sqrt{(k-1)(k+1)}=\sqrt{k^2-1}<k.$$ In your case, you have $k=2$, so $r(2,n)<2+1=3.$