Since $e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac12(1+\frac13(1+\frac14(1+\dots)))$, we have


Is there however a nice way to express the radical in the title too?

  • $\begingroup$ I would try to evaluate it to high precision and run the resulting approximation through a reverse symbolic calculator to see if anything comes up. $\endgroup$ – Bruno Joyal Feb 3 '14 at 22:03
  • $\begingroup$ @BrunoJoyal, I agree it's a good idea, but with one caveat: I ran $4^{e-2}\approx2.70675377667$ through an ISC, just as a "sanity check," and came up empty. $\endgroup$ – Barry Cipra Feb 3 '14 at 22:13
  • $\begingroup$ @BarryCipra Well that's unfortunate... these things aren't perfect. On the other hand, a constant like $4^e$ seems pretty rare, I don't think I've ever gotten anything quite like it when evaluating an integral, say. But it does seem weird that it wouldn't be detected. $\endgroup$ – Bruno Joyal Feb 3 '14 at 22:28
  • $\begingroup$ I have to agree, you probably won't find an answer to this. Noting that the power of the radicals changes (or if the values inside change) makes me want to say there is no closed form. $\endgroup$ – Simply Beautiful Art Feb 1 '16 at 1:43
  • 1
    $\begingroup$ It might be interesting to you that by taking $4$ out of the first root, we obtain another form of this radical: $$2\sqrt{1+\sqrt[3]{\frac{1}{2^4}+\sqrt[4]{\frac{1}{2^{22}}+\dots\sqrt[k]{\frac{1}{2^{k!-2}}+\dots}}}} \approx 2.401615526...$$ You just have to carefully take $1/4$ under each root, moving further and further $\endgroup$ – Yuriy S Jan 16 '17 at 23:10

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