TITLE: Study of the convergence of a sequence with repeated radicals
QUESTION [9 upvotes]: Consider the sequence
$$
a_n = \sqrt {1!\sqrt {2!\cdots\sqrt {n!} } }, \quad n\in\mathbb N.
$$
Does this sequence converge?  
Clearly, $\{a_n\}_{n\in\mathbb N}$ is monotonically increasing.
Therefore, there are two possibilities: 
Either the sequence goes to infinity or it is bounded and therefore, converges to a finite limit.  
Which of the two holds?

REPLY [11 votes]: Note that
\begin{align}
\log a_n&=\log \sqrt{1!\sqrt{2!\cdots\sqrt{n!}}}=\frac{1}{2}\log 1!
+\frac{1}{4}\log 2!+\cdots+\frac{1}{2^n}\log n! \\
&=\sum_{k=1}^n \frac{\log (k!)}{2^k}=\sum_{k=1}^n\frac{1}{2^k}\sum_{j=1}^k\log j=
\sum_{k=1}^n \log k \Big(\sum_{j=k}^n \frac{1}{2^j}\Big).
\end{align}
Therefore, the sequence $\log a_n$, which is increasing, converges to
$$
\log a_n=\sum_{k=1}^n \log k \Big(\sum_{j=k}^n \frac{1}{2^j}\Big)\longrightarrow\sum_{k=1}^\infty\frac{\log k}{2^{k-1}}=b<\infty.
$$
Convergence can be established using for example the ratio test.
Thus
$$
a_n\to \mathrm{e}^b=\exp\left(\sum_{k=1}^\infty\frac{\log k}{2^{k-1}}\right)=\prod_{k=1}^\infty k^{2^{-k+1}}.
$$ 
Note. I am wondering whether $\sum_{k=1}^\infty\frac{\log k}{2^{k-1}}$ can be expressed in terms of some known constants.