Showing that $\sigma=\prod_{n=1}^{\infty}(n!)^{\frac{1}{2^{n+1}}}$

Question

Somos's quadratic recurrence constant

The Somos's Quadratic recurrence constant is defined by the sequence $g_n=ng_{n-1}$ with initial value of $ g_0= 1$

The value of $\sigma=1.661687...$

An infinite product from maths world $\sigma=\prod_{k=1}^{\infty}k^{\frac{1}{2^k}}$

We found another infinite product involving the factorial numbers by experiments on a sum calculator.

$$\sigma=\prod_{n=1}^{\infty}(n!)^{\frac{1}{2^{n+1}}}$$ Where n! is valid for non-negative integers and defined by

$n!=n(n-1)(n-2)\cdots2\cdot1$

Can somebody help us to prove this

Answer 1

This is an opportunity to use summation by parts: $$ \sum _{k=0}^{n} a_kb_k = A_nb_{n}- \sum _{k=0}^{n-1} A_k(b_{k+1}-b_k) \tag1 $$ where $\displaystyle A_n:=\sum _{k=0}^{n} a_k$. Applying it with $$ a_k=\frac1{2^k},\quad A_n=2-\frac1{2^n}, \quad b_k=\log (k!),\quad b_{k+1}-b_k=\log (k+1), $$ gives $$ \begin{align} \sum _{k=0}^{n} \frac1{2^k} \log (k!) &= \left(2-\frac1{2^n}\right)\log (n!)- \sum _{k=0}^{n-1} \left(2-\frac1{2^k}\right)\log (k+1) \\& =\left(2-\frac1{2^n}\right)\log (n!)- \sum _{k=1}^{n} \left(2-\frac2{2^k}\right)\log k \\&=-\frac1{2^n}\log (n!)+2\sum _{k=1}^{n} \frac1{2^k}\log k \end{align} $$ that is

$$ \sum _{k=1}^{n} \frac1{2^k}\log k=\sum _{k=0}^{n} \frac1{2^{k+1}} \log (k!)+\frac1{2^{n+1}}\log (n!) \tag2 $$

By letting $n \to \infty$, using $\displaystyle 0\leq\frac1{2^{n+1}}\log (n!)\leq \frac{n\log n}{2^{n+1}}$, then exponentiating both sides one gets the announced result.

Answer 2

Write $n!=\prod_{k=1}^{n} k$ and you have:

$$\begin{align}\prod_{n=1}^{\infty}(n!)^{\frac{1}{2^{n+1}}}&=\prod_{n=1}^{\infty}\prod_{k=1}^{n} k^{1/2^{n+1}}\\ &=\prod_{k=1}^{\infty}\prod_{n=k}^{\infty}k^{1/2^{n+1}}\\ &=\prod_{k=1}^{\infty}k^{\sum_{n=k}^{\infty}1/2^{n+1}}\\ &=\prod_{k=1}^{\infty}k^{1/2^k}\\ &=\sigma \end{align}$$