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I need help to prove the following inequality with a nested radical.

$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$$

I have tired to use the Stirling approximation, but I got stuck with the nested radical. Does anyone have any idea?

Barry Cipra
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    You can rewrite the nested radical into $1!^{1/2}\cdot 2!^{1/4}\cdots n!^{1/2^n}$. Maybe the Stirling approximation helps now? How about logarithms? – Arthur Jan 04 '18 at 12:40
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    According to my calculation with PARI/GP, the limit for $n\rightarrow \infty$ is $$2.761206841957498033230454647$$ which would contradict the desired inequality. For $n=5$, the expression exceeds $2$ already. – Peter Jan 04 '18 at 12:47
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    BTW, it's almost a duplicate of this question, though in disguise: it's not obvious that both expressions have the same limit. –  Jan 04 '18 at 13:19
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    FWIW, here are a few more digits, calculated using the series in Barry's answer to 250 terms: 2.761206841957498033230454646580131104876125980715304850950745961 And here's the logarithm: 1.015667845736876784378083681444152748492436866865201859073278550 – PM 2Ring Jan 04 '18 at 13:39
  • @ProfessorVector you seem to see duplicate that I don't see –  Jan 05 '18 at 13:35
  • I wrote almost duplicate: the sequences aren't the same, but the limits for $n\to\infty$ coincide. –  Jan 05 '18 at 13:39

3 Answers3

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Following up on Peter's comment below the OP, we have

$$\begin{align} \log\left(\sqrt{1!\sqrt{2!\sqrt{\ldots\sqrt{n!}}}}\right) &={1\over2}\log(1!)+{1\over2^2}\log(2!)+\cdots+{1\over2^n}\log(n!)\\ &=\left({1\over2}+{1\over4}+\cdots+{1\over2^n}\right)\log1+\left({1\over4}+{1\over8}+\cdots+{1\over2^n}\right)\log2+\cdots+{1\over2^n}\log n\\ &\to\log1+{1\over2}\log2+{1\over4}\log3+{1\over8}\log4+\cdots\\ &\gt{1\over2}\log2+{1\over4}\log2+{1\over8}\log2+\cdots=\log2 \end{align}$$

So the requested inequality is not true.

Barry Cipra
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  • Thanks now what if the bound is 3 instead of 2? –  Jan 04 '18 at 13:09
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    You should ask about $3$ as a new question, with a link to this one for context, rather than adding it here. (I've rolled back to the version without the $3$.) – Barry Cipra Jan 04 '18 at 13:15
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We have that

$$\sqrt{1!\sqrt{2!\sqrt{\cdots\sqrt{\text{n}!}}}}=\prod_{k=1}^n(k!)^{1/2^k}=\prod_{1\le j\le k\le n}j^{1/2^k}=\prod_{j=1}^nj^{\sum_{k=j}^n1/2^k}=\prod_{j=1}^nj^{(1/2)^{j-1}-(1/2)^n}\\=\exp\left(\sum_{j=1}^n\ln j\left(\frac1{2^{j-1}}-\frac1{2^n}\right)\right)>\exp\left(\left(\sum_{j=2}^n\left(\frac12\right)^{j-1}\right)-\frac{n-1}{2^n}\right)\to\exp(1-0)=e$$

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You might consider this.
$$\sqrt{1!\sqrt{2!\sqrt{\cdots\sqrt{\text{n}!}}}}=\left(1!\right)^\frac{1}{2}\times\left(2!\right)^\frac{1}{4}\times\cdots\times\left(\text{n}!\right)^\frac{1}{2^\text{n}}=\exp\left\{\sum_{\text{k}=1}^\text{n}\frac{1}{2^\text{k}}\cdot\ln\left(\text{k}!\right)\right\}\tag1$$

Can you bound the sum inside. It is convergent when $\text{n}\to\infty$?

$$\lim_{\text{k}\to\infty}\left|\frac{\frac{1}{2^{\text{k}+1}}\cdot\ln\left(\left(\text{k}+1\right)!\right)}{\frac{1}{2^\text{k}}\cdot\ln\left(\text{k}!\right)}\right|=\frac{1}{2}<1\tag2$$

Jan Eerland
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