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Everyone knows about the classic $$ \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 $$ However, is there any way to find $$ \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 + \dfrac14 + \dfrac{1}{16} + \dfrac{1}{256} + \cdots $$

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  • $\begingroup$ okay i edited the post $\endgroup$ – Akshaj Kadaveru Nov 27 '13 at 17:34
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    $\begingroup$ In fact, it's transcendental (look for Fredholm number). As such, it's unlikely to have any "nicer" expression, just as @DanShved suggested. $\endgroup$ – Peter Košinár Nov 27 '13 at 17:41
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    $\begingroup$ math.stackexchange.com/questions/276892/… $\endgroup$ – Mats Granvik Nov 27 '13 at 17:43
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    $\begingroup$ A fairly common keyword is "lacunary series", or "gap series" to google for. It has been a nice exercise (I think) in the beginning of the 20'th century to prove transcendentality of that number ... $\endgroup$ – Gottfried Helms Feb 10 '17 at 23:43
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    $\begingroup$ Perhaps an interesting additional input about generalizations and especially the alternating gap/lacunary series go.helms-net.de/math/divers/mo/MO_Lacunaryseries.pdf $\endgroup$ – Gottfried Helms Aug 25 '17 at 19:20
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This is the Fredholm number as pointed out in the comments, but in case you are interested in the numerical value, a quick Mathematica calculation reveals the sum is approximately $0.816422$.

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The sum mentioned by the OP, and sums like it, are in general very difficult to deal with. The OP's series is actually called the Fredholm number, and is transcendental (see the Wikipedia page for Transcendental number).

Note: This constant is also known as Kempner's number (not to be confused with the Kempner series!).

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$$\sum_{k=0}^{\infty}\frac{1}{a^{a^k}}=\sum_{k=0}^{\infty}a^{-a^k}=\sum_{k=0}^{\infty}e^{-a^k\log(a)}$$ Let's choose the seemingly simple example $a=e$ to get rid of the logarithm. We then get the sum $$\sum_{k=0}^{\infty}e^{-e^k}$$
Now, Wolfram Alpha can't tackle it, nor can Mathematica. Given that this is a simple example of what this sort of sum looks like, I'm betting that no known, simple closed form exists. If I'm wrong, feel free to correct me though. I would be pleasantly surprised.

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