Problem: Let $$f(x) = \sum_{k\ge0}(1-e^{-x/2^k}).$$ Give an asymptotic expression for $f$.
According to this post, we have: $$ f(x) \underset{x\to+\infty}{=} \log_2(x) + \mathrm{O}\left(\ln\ln x\right)$$
I'd like to know if we can solve this problem with my solution so far, and I'd be glad if you could give some hints on what to do next:
Notice that: $$ f(2x)-f(x) = \sum_{k\geq 0} \left(\exp\left(\dfrac{-x}{2^k}\right)-\exp\left(\dfrac{-x}{2^{k-1}}\right)\right) = 1-\exp(-2x) = 1 + o(1)$$ Since $\log_2(x)$ is a solution of $f(2x)-f(x) = 1$, we'll introduce $g(x)=f(x)-\log_2(x)$. According to the expected solution, we need to show that : $$ g(x) = O(\ln \ln x)$$
This is where I'm stuck, but these are the results I found so far: $$ g(2x) - g(x) = f(2x)-f(x)-\log_2(2x)+\log_2(x) = \left(1-\exp(-2x)\right)-1 = -\exp(-2x)$$ $\bullet$ Therefore, for $n\geq 1$, $$ g(2^nx)-g(x) = -\sum_{k=0}^{n-1} \exp(-2^kx)$$ For $x=1$, $ g(2^n) \leq g(1) = \displaystyle\sum_{k\geq 0} \left(1-\exp\left(\dfrac{-1}{2^k}\right)\right)$, and I can prove that this series is convergent
$\bullet$ Let $h(x) = \sum_{k\geq 1} \exp(-2^kx)$. For $x\geq 1$, we have $h(x) = O(1)$. We also have : $$ h(2x)-h(x) = -\exp(-2x) = g(2x)-g(x)$$ So for $\phi(x) = g(x)-h(x)$, we have : $$ \phi(x) = \phi(2x)$$
This is not a full answer.
$$I(x)=\int_0^\infty\Bigg( 1-\exp\left(-\frac x {2^k} \right)\Bigg)\,dk$$ $$I(x)=\frac{\log (x)+\gamma }{\log (2)}+\frac{\text{Shi}(x)-\text{Chi}(x)}{\log (2)}$$ where appear the hyperbolic sine and cosine integrals.
When $x$ is large $$\text{Shi}(x)-\text{Chi}(x)=\frac {e^{-x}} x \sum_{n=0}^\infty (-1)^n \,\frac {n!}{x^n} \sim \frac {e^{-x}} x $$ and, using Euler-MacLaurin formula, the result already given by @Svyatoslav in the linked post.