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.
 A: 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.
A: After some analysis I was unable to find a closed form, but here are some good approximations:
On $n\in(0,7)$, $f(n)$ is close to the form of $\alpha + \beta \sqrt n + \gamma n$, and for $n\in[7,\infty)$, is of the form $\lambda + \rho n + \chi \log n$. So we can approximate with,
$$ f(n) \approx \begin{cases}
 \alpha + \beta \sqrt n + \gamma n, & n\in(0,7) \\
 \lambda + \rho n + \chi \log n, & n\in[7,\infty) \\
\end{cases} $$
You can use any fitting algorithm to approximate these values. I came up with,
\begin{align*}
 \alpha &= -3.03894 &\lambda &= -3.23294 \\
 \beta &= 2.17427  &\rho &= 1.04346 \\
 \gamma &= 0.868928  &\chi &= 2.28405 \\
\end{align*}
If computing a logarithm is too expensive or computation heavy, note that $f$ is approximately linear as $n$ grows. You can approximate $f(n)\approx 1.09352 n + 2.82223$.
