I am trying to find the Mellin transform of the function
$$
G(z) = \sum_{k \ge 1} C_k\left( 1- \exp \left( \frac{-z}{4^k} \right )\right),
$$
where $C_k$ denotes the $k$-th Catalan number ($C_k = \frac 1{n+1}\binom{2n}n $), yet the function doesn't seem to have a transform.
Follwing the paper Mellin Transforms and Asymtotics: Harmonic Sums (mainly Section 3) by Philippe Flajolet et al, we first separate the Dirichlet series from the harmonic sum, and, in order to stay consistent with their treatment, switch $G(z)$ for $G(1/z)$. In their notation, using
$$
\lambda_k = C_k,\, \mu_k = 4^k,\, g(z) = 1-\exp(-1/z)
$$
we get the Dirichlet series
$$
\Lambda(s) = \sum_{k \ge 1 } C_k 4^{-ks} \approx \sum_{k \ge 1 } \frac 1 {\sqrt{ \pi k^3}}4^{k(1-s)}.
$$
The problem now is that $\Lambda(s)$ converges for $\mathrm{Re}(z) > 1$, and the base function $g(z)$ has the fundamental strip of $(0,1)$, meaning that we can't factor the series from the base function following their Lemma 2, nor can we read of the asymptotics of $G(z)$ by using the singular expansion of $G^*(z)$ (which is what I would like to do).
My question is hence:
Is there a different way to obtain the Mellin transform of $G(z)$?
The asymptotic I expect is $G(z) \sim \frac{z}{\sqrt{ \log z}}$, for $z \to \infty$, which an asymptotic I have not found by any Mellin approach so far. Nor do the translation theorems in Flajolet's paper really seem to cover such a function. So maybe that's the problem.
PS The last part might be a question on its own respect, but it might cover the same issue. I might want to ask the second part as an own question.