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let $n\in N^{+}$ and such $n\ge 2$

prove or disprove $$\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}\tag{1}$$

and I have use induction prove $$\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<3$$ because $$\sqrt{(k+1)(k-1)}<k$$ and we known this Ramanujan's indentity: Prove $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3$ and also use indution prove this $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}<3$$ see:Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $.

But how to prove $(1)$

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1 Answers1

1

$$L=\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}$$ $$\log L=\sum_{k=1}^{\infty }\frac{\log(k+1)}{2^k}=\quad{\text{LerchPhi}^{0,1,0}}(\frac{1}{2},0,2)=1.0156678457368767......$$ so $$L=e^{1.01566784573687........}=2.761206841957498033230454646......$$ hence $$2.7612.....<3$$

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