Is it possible to find the point of convergence of $\prod_1^\infty k^{\frac1{k!}}$


My attempt:

If $S_n=\prod_1^\infty k^{\frac1{k!}}$ then $\ln S_n=\sum_1^\infty \frac{\ln k}{k!}< \sum\frac1{k!}=e$

So $S_n$ is converges but I don't know if it is possible to find the point of convergence or not.

  • $\begingroup$ But $\log k < 1$ only for $k=1,2$, no? $\endgroup$ – krvolok Aug 10 '15 at 16:53
  • $\begingroup$ @krvolok Yeah sorry I should say $\sum\frac{\ln k}{k!}< \sum\frac k{k!}=\sum\frac1{(k-1)!}$ $\endgroup$ – mac Aug 10 '15 at 16:55
  • $\begingroup$ The value of the product is $ 1.829024680$. $\endgroup$ – user64494 Aug 10 '15 at 16:57
  • $\begingroup$ @user64494 Why!? $\endgroup$ – mac Aug 10 '15 at 18:27
  • $\begingroup$ @ mac : That was found by the Maple code $$>evalf(product(k^{1/factorial(k)}, k = 1 .. infinity));$$ $\endgroup$ – user64494 Aug 10 '15 at 19:37

It's possible to rephrase your product as a nested radical, and sometimes these have closed-form solutions, like $\sqrt{2 \sqrt {2 \sqrt {2 \ldots}}} = 2$. However no one even knows a closed-form solution for the seemingly almost as innocent $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$. There all the nested radicals are square-roots. Your nested radicals are $k$th roots for increasing $k$. Thus it seems highly unlikely that anyone will be able to come up with a closed form for your expression, at least, not without some monumental effort.


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