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Find a closed form for $\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^{2^2}}+\frac{1}{2^{2^{2^2}}}+\cdots $

Since I've never encountered this type of series before I was hoping someone here could help me start. Thanks!

EDIT: I've edited some stuff in light of the comments. I have also found an interesting manipulation, but I don't think it leads anywhere seeing the comments. Here it is:

Let $S$ be the series. Define $f(x)=\underbrace{2^{2^{2^{\cdots}}}}_{k\ 2's}+\underbrace{2^{2^{2^{\cdots}}}}_{k+1\ 2's}$, e.g. $f(2)=2^2+2^{2^2}$. Then adding pairs of terms we can write $S$ as:

$S=\frac{f(1)}{2^{f(0)+1}}+\frac{f(2)}{2^{f(1)+1}}+\frac{f(3)}{2^{f(2)+1}}\cdots$

$=\frac{f(1) 2^{f(1)-1}}{2^{f(0)+f(1)}}+\frac{f(2) 2^{f(2)-1}}{2^{f(1)+f(2)}}+\frac{f(3) 2^{f(3)-1}}{2^{f(2)+f(3)}}+\cdots $

And then you can use $x 2^{x-1}+\sum_{k=1}^x k\binom{x}{k}$?

Marc van Leeuwen
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  • Extremely unlikely there's a closed form, I think. Do you have some reason to believe there is one? Also, if you look up hypergeometric functions, I think you'll see this has nothing to do with them. – Gerry Myerson Sep 12 '14 at 12:59
  • Just by looking at it's base two representation it's easy to see that the number is irrational. Other then that I doubt you will find anything someone would consider a "closed form" representation for this series. – Ethan Splaver Sep 12 '14 at 13:06
  • @EulerianAdventurer Now that I revisit, mine was correct I believe. $a_1 = \frac12, a_{n+1} = a_n^2$. We are looking for $$\sum_{n=1}^\infty a_n < 2 - \frac18$$ Convergence is obvious and $<2 - \frac18 - \ldots$ as well. – AlexR Sep 12 '14 at 13:09
  • This is not a (generalised) hypergeometric series. Was the typo in the title intentional? – Marc van Leeuwen Sep 12 '14 at 13:09
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    This is the same as the series considered in this question, evaluated at $x=1/2$. The commentary on this answer on MathOverflow may also be interesting; apparently, it can be shown that your series is not just irrational but moreover transcendental. (That tells you nothing about a closed-form, of course!) – Semiclassical Sep 12 '14 at 13:42

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