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

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?

• Did you check Masacroso's answer? It does prove the new identity ($e<3$). Commented Jan 5, 2018 at 13:51
• Actually, the values of this is $2.7612068419574980332304546465801311048761259807153048509507‌​45961$, so even less than $\frac{70}{25}$. Commented Jan 5, 2018 at 13:52
• @DietrichBurde thanks but in my numerics I need the smallest integer. that is why I persist with 2 or 3
– user518372
Commented Jan 5, 2018 at 13:55
• @DietrichBurde my bad! When you skim too fast. Thanks anyway. Commented Jan 5, 2018 at 13:56
• The answer given here (math.stackexchange.com/questions/472945/…) as $n\to\infty$ solves your question completely because your series given in the answer here (math.stackexchange.com/questions/2591546/…) is the same series. Commented Jan 5, 2018 at 14:55

@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.

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.

• THis inequality is somehow surprising. How the term with factorial be less than the other?
– user518372
Commented Jan 8, 2018 at 16:02
• I am pretty sure that we have$$\sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}} < \sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}}$$ not the other way around.
– user518372
Commented Jan 8, 2018 at 16:03
• @Sobolev Why do you think so? Which part of my proof is incorrect or unclear to you? I'll be glad to add some more explanation. Commented Jan 8, 2018 at 17:52
• You should suppose $n\geq 2$. Commented Sep 29, 2021 at 8:43

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.

• amazing... I tried to find some analytic bound but I failed miserably. Congrats!
– user173262
Commented Jan 5, 2018 at 18:30

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 $4$.