Is the sum of series $\displaystyle \sum_{n=0}^\infty \frac1{2^{2^n}} = \frac12 + \frac14 + \frac1{16} + \frac1{256} + \frac1{65536} + \dotsb \approx 0.8164215090218931$ representable in a closed form? If so, what it is?
1 Answer
If you allow for a named number to be a closed form representation, the answer is "yes".
$\sum_{n=0}^\infty (1/2)^{2^n}$ is known as the Kempner number [1], a transcendental number [2].
More generally, numbers of the type $\sum_{n=0}^\infty x^{2^n}$ with algebraic $x\in\mathbb{C}$, $x<1$, are known to be transcendental, as first proven by Mahler [3].
[1] A.J. Kempner, On Transcendental Numbers (1916).
[2] Boris Adamczewski, The many faces of the Kempner number (2013).
[3] K. Mahler, Arithmetische Eigenschaften einer Klasse transzendentaltranszendenter Funktionen (1930).

2$\begingroup$ Since 2 is rational, doesn't this also just follow from Liouville's theorem? Mahler's theorem is interesting precisely because it includes irrational algebraic numbers. $\endgroup$– JoshuaZCommented Jul 16 at 11:30

1$\begingroup$ if I understand Adamczewski correctly, Liouville needs a faster than exponential growth of the exponent (so $n!$ rather than $2^n$) $\endgroup$ Commented Jul 16 at 11:37

2$\begingroup$ Huh. For some reason I thought exponential was sufficient. $\endgroup$– JoshuaZCommented Jul 16 at 14:59