# Compute $\sum_{k=0}^{\infty}\frac{1}{2^{k!}}$

How could the series below be computed ? $$\sum_{k=0}^{\infty}\frac{1}{2^{k!}}$$ It's not a series from a book, but a series I thought of many times, and I didn't
manage to figure out what I should do here. I'm just curious to know if there are
some known ways for approaching such a series. Thanks!

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I've reverted your edit; I think changing it to just "Compute [the sum]" might attract down-votes, since questions that take the form of commands tend to be frowned upon by some here. – Clive Newstead Jan 19 '13 at 13:52
@CliveNewstead: thanks for your concern! After a while I prefer to delete the explanations I give when I firstly post a question and only want to let the essential part. I think the questions look better with a concise appearance. :). OK, I let this question the way it is. – user 1618033 Jan 19 '13 at 14:04

## 3 Answers

This is a binary version of Liouville's constant, one of the first explicitly given numbers to be proved transcendental. I don't think it has any nicer expression than the series itself.

Of course it is easy to write down the sum, as a binary fraction!

$$1.01000100000000000000000100000..._2$$

and it is also easy to produce the decimal representation digit by digit -- except for the first few terms, the first significant digit of each term comes after the last nonzero digit of the terms before, so we get

$$1.265625059604644775390625000000..._{10}$$

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(+1) Interesting. This is new to me. Thanks for your answer! – user 1618033 Jan 15 '13 at 22:33
I might use this number in my discrete dynamical systems class. It gets closer and closer to zero under the doubling mapping but never quite gets there! – Jp McCarthy Jan 15 '13 at 22:35

In fact, your series very likely has no better closed-form than that, and its value is transcendental. This can be explained by the fact that the number is a so-called Liouville Number — in a sense, it's 'too close' to its rational approximations to be an algebraic number. For more details on a closely-related number (replacing your $2$ by $10$, have a look at http://mathworld.wolfram.com/LiouvillesConstant.html (and Liouville's number revisited ) — in particular, it turns out that the continued fractions for these numbers have a surprisingly explicit form.

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Thanks for your answer! (+1) – user 1618033 Jan 15 '13 at 22:34

I very much doubt it can be given in a closed form, but if your interested in very fast converging series that can be given in closed form here are a couple:

$$\sum_{n=0}^\infty \frac{2^n}{x^{2^n}+1}=\frac{1}{x-1}$$ $$\sum_{n=0}^\infty \frac{3^n(3^{3^n}+2)}{(x^2)^{3^n}+x^{3^n}+1}=\frac{1}{x-1}$$ $$\int_{0}^1\frac{1}{x^x} \ dx=\sum_{n=1}^\infty\frac{1}{n^n}$$

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thanks. It might help in my future studies. (+1) – user 1618033 Jan 15 '13 at 22:43
@Ethan, do you have any online references, or outline for proofs of these identities? – user45099 Jan 15 '13 at 22:49
The first two I found from playing with geometric series, the third one is called the Sophmore's dream it can be found here en.wikipedia.org/wiki/Sophomore's_dream. – Ethan Jan 15 '13 at 22:52
But that has absolutely nothing to do with the question! – Pedro Tamaroff Jan 15 '13 at 23:51