# Is $\sqrt{2\sqrt[3]{3!\sqrt[4]{4!\sqrt[5]{5!..}}}}=2\log \pi$

Is $$\sqrt{2\sqrt[3]{3!\sqrt[4]{4!\sqrt[5]{5!..}}}}=2\log \pi$$

Can anyone help me to know the way of proving above if it is true?

• Trying taking log of the expression on the left side. – Singhal Aug 12 '16 at 6:49
• @Singhal I think that this not easy – SADIK MOHAMMED Aug 12 '16 at 6:50
• It may help to write out the expression as $\prod_n (n!)^{1/n!}$. From there you could get an approximation using Stirling's, but that would only be an approximation. I'm not sure how to get an exact value. – Aaron Aug 12 '16 at 6:57
• Why did you tell us to prove something that's not true? – arctic tern Aug 12 '16 at 7:05
• @arctictern I did not know that it is not true, sorry – SADIK MOHAMMED Aug 12 '16 at 7:12

They are not equivalent: If we define $$R_n = (2!(3!(\cdots(n!)^{1/n}\cdots)^{1/3})^{1/2},$$ then $$R_{20} \approx 2.2902182705436512846, \\ R_{200} \approx 2.2902182705436512868,$$ which suggests that the convergence is at least accurate to $10^{-3}$; but $$2 \log \pi \approx 2.2894597716988003483.$$

Note we can also write the expression in a non-nested fashion: $$R_n = \prod_{k=2}^n (k!)^{1/k!} = \exp \left( \sum_{k=2}^\infty \frac{1}{k!} \log k! \right).$$ This sum converges extremely rapidly; only a few terms are needed to obtain the necessary precision to prove inequivalence.

We can proceed further: \begin{align*} \log R_n &= \sum_{k=1}^\infty \frac{1}{k!} \sum_{j=1}^k \log j \\ &= \sum_{j=1}^\infty \log j \sum_{k=j}^\infty \frac{1}{k!} \\ &= \sum_{j=1}^\infty \frac{e \log j}{\Gamma(j)} \int_{t=0}^1 t^{j-1} e^{-t} \, dt \\ &= e \int_{t=0}^1 e^{-t} \sum_{j=0}^\infty \frac{\log (j+1)}{j!} t^j \, dt. \end{align*} However, at this point, I'm not sure how to evaluate this sum, which clearly converges on $t \in [0,1]$. It seems as if it should be well-known; it is a Taylor series of a function $f(t)$ whose $j^{\rm th}$ derivative evaluated at $t = 0$ equals $\log (j+1)$.

• Thanks(+1) but I ask if there is a closed form. – SADIK MOHAMMED Aug 12 '16 at 7:03
• @SADIKMOHAMMED: What is a closed form? – Fabian Aug 12 '16 at 7:04
• @Fabian closed form of the L.H.S – SADIK MOHAMMED Aug 12 '16 at 7:05
• (+1) I didn't see your edit before I answered. – robjohn Aug 12 '16 at 7:06
• @SADIKMOHAMMED Actually, you did not ask if there is a closed form. You asked for a proof for a relationship that is not true, and the response you have received thus far has addressed that. – heropup Aug 12 '16 at 7:17

\begin{align} \log\left(\sqrt{2\sqrt[3]{3!\sqrt[4]{4!\sqrt[5]{5!..}}}}\right) &=\log(2)\left(\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+\dots\right)\\ &+\log(3)\left(\frac1{3!}+\frac1{4!}+\frac1{5!}+\dots\right)\\ &+\log(4)\left(\frac1{4!}+\frac1{5!}+\dots\right)\\ &+\log(5)\left(\frac1{5!}+\dots\right)\\[6pt] &+\dots\\[6pt] &=\sum_{k=2}^\infty\frac{\log(k!)}{k!} \end{align} Summing to $k=22$ gives $20$ places of accuracy, and that gives $$\sqrt{2\sqrt[3]{3!\sqrt[4]{4!\sqrt[5]{5!..}}}}=2.2902182705436512868$$ Since $2\log(\pi)=2.2894597716988003483$, the identity is false.

• Would the downvoter care to comment? – robjohn Aug 14 '16 at 12:08

Note that your expression can be written as $$\prod_{n=2}^\infty (n!)^{1/n!}.$$If you're not willing to compute that many partial sums, we can see it is larger than $2\log\pi$ because \begin{align}\sum_{n=2}^\infty \frac{\log n!}{n!}>\frac{\log2}{2}+\frac{\log6}{6}+\sum_{n=4}^{\infty}\frac{n-1}{n!}&=\frac{\log2}{2}+\frac{\log6}{6}+\sum_{n=3}^\infty\frac1{n!}-\frac1{(n+1)!}\\ &=\frac{2+4\log2+\log3}{6}>\log2+\log\log\pi,\end{align}where we have used $\log n!>n-1$ for $n\ge4$, easy to show by induction.