# Sum of Infinite Series $1 + 1/2 + 1/4 + 1/16 + \cdots$

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

• okay i edited the post – Akshaj Kadaveru Nov 27 '13 at 17:34
• In fact, it's transcendental (look for Fredholm number). As such, it's unlikely to have any "nicer" expression, just as @DanShved suggested. – Peter Košinár Nov 27 '13 at 17:41
• math.stackexchange.com/questions/276892/… – Mats Granvik Nov 27 '13 at 17:43
• 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 ... – Gottfried Helms Feb 10 '17 at 23:43
• Perhaps an interesting additional input about generalizations and especially the alternating gap/lacunary series go.helms-net.de/math/divers/mo/MO_Lacunaryseries.pdf – Gottfried Helms Aug 25 '17 at 19:20

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 $0.816422$.
$$\sum_{k=0}^{\infty}\frac{1}{a^{a^k}}=\sum_{k=0}^{\infty}a^{-a^k}=\sum_{k=0}^{\infty}e^{-a^k\log(a)}$$ Let's choose the seemingly simple example $a=e$ to get rid of the logarithm. We then get the sum $$\sum_{k=0}^{\infty}e^{-e^k}$$