How to prove :$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$

Question

In fact initially I wanted to prove that $$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$$

Which by the accepted answer here fails to be true.

@Barry Cipra advised me to ask this question in a different post: How can I now prove that: $$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$$

Note that the answers here do not provide an estimate of this sequence

Can anyone have any idea?

Answer 1

@BarryCipra has provided this limit here: $$\log\left(\sqrt{1!\sqrt{2!\sqrt{\ldots\sqrt{n!}}}}\right)\to\log1+{1\over2}\log2+{1\over4}\log3+{1\over8}\log4+\cdots$$ We can write the RHS as $$\frac12\log2+\frac14\log2+\frac14\log\frac32+\frac18\log2+\frac18\log\frac42+...$$ or $$\log2+\frac14\sum_{n=0}^\infty\frac1{2^n}\log\left(\frac{n+3}2\right)$$ since $\frac12+\frac14+\frac18+...=1$. Now as $$\log\left(\frac{n+3}2\right)<1.2^{n-2.39}$$ by Desmos, $$\text{RHS}<\log2+\frac14\sum_{n=0}^\infty\frac{1.2^{n-2.39}}{2^n}=\log2+\frac1{4\cdot1.2^{2.39}}\sum_{n=0}^\infty0.6^n$$ and by the geometric series with $a=1$ and $r=0.6$, $$\text{RHS}<\log2+\frac1{4\cdot1.2^{2.39}}\cdot2.5=1.097...$$ Hence $$\boxed{\sqrt{1!\sqrt{2!\sqrt{\ldots\sqrt{n!}}}}<e^{1.097...}=2.996...<3}$$ as desired.

Answer 2

For $n\geq 2$, \begin{align*} \sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} & = 2^{1/4+1/8+1/16+\ldots+1/2^n} \cdot 3^{1/8+1/16+1/32+\ldots+1/2^n}\cdot \ldots \cdot n^{1/2^n} \\ & < 2^{1/2}\cdot 3^{1/4} \cdot 4^{1/8} \cdot \ldots \cdot n^{1/2^{n-1}} \\ & = \sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}} \\ &<3\end{align*} The last inequality was proved here.

Answer 3

For any $n > 0$, we have

$$s_n \stackrel{def}{=} \log\left(\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}}\right) = \sum_{k=1}^n \frac{\log(k!)}{2^k} = \sum_{k=1}^n \frac{1}{2^k}\sum_{j=1}^k\log(j) $$ Recall for any function $f(z)$ with power series representation $\sum\limits_{k=0}^\infty a_k z^k$,, the function $\frac{f(z)}{1-z}$ has the power series representation $\sum\limits_{k=0}^\infty z^k \sum\limits_{j=0}^k a_j$. Apply this to the limit of $s_n$, we have

$$\begin{align} s_\infty \stackrel{def}{=} \lim_{n\to\infty} s_n &= \frac{1}{1 - \frac12} \sum_{k=1}^\infty \frac{\log k}{2^k} = \sum_{k=1}^\infty\frac{\log(k+1)}{2^k}\\ &= \sum_{k=1}^\infty\frac{1}{2^k}\sum_{j=1}^k(\log(j+1)-\log(j)) = \frac{1}{1 - \frac12} \sum_{k=1}^\infty\frac{\log(k+1)-\log(k)}{2^k}\\ &= \sum_{k=1}^\infty \frac{1}{2^{k-1}}\log\left(1+\frac1k\right)\\ &= \sum_{k=1}^\infty \frac{1}{k2^{k-1}} + \sum_{k=1}^\infty \frac{1}{2^{k-1}}\left[\log\left(1+\frac1k\right) - \frac1k\right]\\ &= 2\log(2) + \sum_{k=1}^\infty \frac{1}{2^{k-1}}\left[\log\left(1+\frac1k\right) - \frac1k\right] \end{align} $$ Notice $\log(1+x) \le x$ for all $x \in (-1,\infty)$, we can truncate the second series at some finite $p$ and turn it to an upper bound $$ s_\infty \le 2\log(2) + \sum_{k=1}^p \frac{1}{2^{k-1}}\left[\log\left(1+\frac1k\right) - \frac1k\right]$$ Take $p = 1$, this becomes $s_\infty \le 3\log(2) - 1$. As a result,

$$\lim_{n\to\infty} \sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} = \lim_{n\to\infty} e^{s_n} = e^{s_\infty} \le e^{3\log(2)-1} = \frac{8}{e} \approx 2.943035529371539 < 3$$

We can improve the bound by using a larger $p$. For example, if we take $p = 3$, we obtain a new bound $e^{s_\infty} \le 2.775306746902055$. This is within 1% of the correct limit $\approx 2.761206841957498$ pointed out by others in comment.

Answer 4

SKETCH FOR AN ANALYTIC APPROXIMATION TO THE BOUND (too long for a comment):

From my previous answer we have that

$$S_n:=\sqrt{1!\sqrt{2!\sqrt{\cdots\sqrt{\text{n}!}}}}=\exp\left(\sum_{j=1}^n\ln j\left(\frac1{2^{j-1}}-\frac1{2^n}\right)\right)\\=\exp\left(\left(\sum_{j=0}^{n-1}\frac{\ln (1+j)}{2^j}\right)-\frac{\ln(n!)}{2^n}\right)$$

And because $0\le\ln(n!)/2^n\le \ln(n^n)/2^n\le n^2/2^n$ then

$$S:=\lim_{n\to\infty} S_n=\exp\left(\sum_{k\ge 0}\frac{\ln (k+1)}{2^k}\right)$$

Now using summation by parts and taking $\Delta g_k=(1/2)^k$ and $f_k=\ln(1+k)$ we find that $g_k=-(1/2)^{k-1}$, thus

$$\sum_{k\ge 0}\frac{\ln(1+k)}{2^k}=\left[-\frac{\ln(1+k)}{2^{k-1}}\right]_{k=0}^{k=\infty}+\sum_{k\ge 0}\left(\frac12\right)^k\ln\left(1+\frac1{k+1}\right)\\=\sum_{k\ge 0}\left(\frac12\right)^k\ln\left(1+\frac1{k+1}\right)=\sum_{k\ge 0}\sum_{j\ge 1}\left(\frac12\right)^k(-1)^{j+1}\frac1{j(k+1)^j}\\\le\sum_{j= 1}^{2n+1}\frac{(-1)^{j+1}}{j}\sum_{k\ge 0}\left(\frac12\right)^k\left(\frac1{1+k}\right)^j=2\sum_{j=1}^{2n+1}\frac{(-1)^{j+1}}j{\rm Li}_j(1/2)$$

for all $\in\Bbb N$, where ${\rm Li}_j$ is a polylogarithm. Thus taking $n=0$ I found the upper bound of $4$.