# Study of the convergence of a sequence with repeated radicals

Let the sequence $$a_n = \sqrt {1!\sqrt {2!\cdots\sqrt {n!} } }$$ Does this sequence converge?

I can tell intuitively that $a_n$ 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.

How can show it?

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Interesting Problem! –  Hawk Feb 24 at 12:05
Does Stirling's Aprroximation work here? –  Hawk Feb 24 at 12:11
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.