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!
 A: 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
$$ 1.265625059604644775390625000000..._{10} $$
A: 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.  
A: 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:
$$\sum_{n=0}^\infty \frac{2^n}{x^{2^n}+1}=\frac{1}{x-1}$$
$$\sum_{n=0}^\infty \frac{3^n(3^{3^n}+2)}{(x^2)^{3^n}+x^{3^n}+1}=\frac{1}{x-1}$$
$$\int_{0}^1\frac{1}{x^x} \ dx=\sum_{n=1}^\infty\frac{1}{n^n}$$
