I know that for a sequence $(p_n)_n$ with $\sum_{i\geq 1} p_n=\infty$ there is a strong relation between the asymptotic behaviour of $p_n$ (or at least $\sum_{k=1}^n p_k$) for $n\to\infty$ and the asymptotic behaviour of the Laplace transform $\mathfrak L_{ (p_n)_n}(\lambda)$ for $\lambda\rightarrow 0$. At least for $p_n$ decaying like $n^{\alpha-1}$ for some $\alpha>0$ one has good asymptotics by some Tauberian Theorems (see i.e. Feller Vol. 2).
Now let's think of a probability measure on $\mathbb N$ which of course satisfies $\sum_{n\geq1} p_n =1$ and also $\sum_{n\geq 1} n p_n = \infty$. One could now compute at least the asymptotics of $$1- \mathfrak L_{(p_n)_n} (\lambda), \quad \lambda \rightarrow 0. \quad (*)$$
But my question is the other way round. Assume we know the asymptotic of $(*)$ and want to know something about the asymptotic behaviour of $p_n$ for $n\rightarrow \infty$. Is there any relation? At least if we have that for some constant $C>0$ and $\alpha>0$ $$1- \mathfrak L_{(p_n)_n} (\lambda)\sim C \lambda^{\alpha}, \quad \lambda \rightarrow 0$$ holds? Maybe that for some constant $\tilde C$ $$1-\sum_{k=1}^n p_k = n^{-\alpha} \tilde C$$ holds?
All the best