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Is there a closed-form expression for $f(x) = \sum_{k=0}^\infty x^{(2^k)}$ (for $|x| < 1$)? Does this function have a name?

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    It is a known pathological example in calculus and a solution of a certain functional equation. No, it can't be expressed in elementary functions, and I don't think it has a name. – 2017-01-16
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    @IvanNeretin good answer. thanks. – 2017-01-16
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    @IvanNeretin What is the functional equation? – 2017-01-16
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    @Arthur Plug $x^2$ as a variable and see how the resulting expression is _almost_ similar to $f(x)$. – 2017-01-16

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It's a lacunary function. It doesn't have a closed form, and interestingly enough it has a singularity at every point on the unit circle, so it can't be analytically continued outside of the unit circle.

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    I'd rather point out that it is merely _one of many_ lacunary functions. – 2017-01-16
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There seems to be no closed form expression for the sum in terms of known functions.

It might be of interest, however, to have some simple bounds for the sum in terms of closed form expressions. We derive these by replacing the sum by an integral.

Let

$$f(x) = \sum _{n=0}^{\infty } x^{2^n}$$

and

$$g(x)=\int_0^{\infty } x^{2^t} \, dt = -\frac{\text{Ei}(\log (x))}{\log (2)} $$

It is then easy to see by comparing the graphs of the summands and the integrand that

$$g(x) < f(x) < x + g(x)$$

A fairly good approximation to $f$ is then the arithmetic mean of upper and lower bound

$$ f(x) \simeq g(x) + x/2$$

Close to $x = 1$ we have asymptotically

$$g(x\to 1) \simeq \frac{\frac{1-x}{2}-\log (1-x)+\gamma }{\log (2)}$$

That is, the sum has a logarithmic divergence for $x \to 1$.

Remark: the discussion is also valid and the formulas hold if the number $2$ is replaced by any real number $a > 1$