Closed form for $\sum_{0\le x\lt\infty}n^{2^{-x}}-1$ Is there a closed form for the expression below?
$$f(n)=\sum_{0\le x\lt\infty}n^{2^{-x}}-1=(n-1)+(\sqrt{n}-1)+(\sqrt{\sqrt{n}}-1)+\cdots$$
Approximations are good as well. This appeared to me while analyzing an algorithm.

Let $g(x) := f(e^{x/2})$ so that $g(x) = g(x/2) + e^{x/2}-1$. Expanding $g(x)$ in power series we have
$$ g(x) = \sum_{k>0} \frac{x^k}{k!(2^k-1)} = 
x + \frac{x^2}{6} + \frac{x^3}{42} + \frac{x^4}{360} + \frac{x^5}{3720} +\cdots $$ which converges everywhere but is unlikely to have closed form.
Approximations of $f(x)$ around $x=1$ where $f(1)=0$ are $x-1+\log(x)$ and $x-3+2\sqrt{x}$ and the average of the two is even better.
