I would like to compute that serie if it is possible : $\sum_{j\geq 0}\frac{(\lambda m^{-1/2})^j}{j!} \exp{\left(-B\,m^{-j}\right)}$ where $B \geq 0,\, \lambda > 0,\, m>1$.
I know it converge as for all $j\geq 0$, $\exp{\left(-B\,m^{-j}\right)}\leq1$ and $\sum_{j\geq 0}\frac{(\lambda m^{-1/2})^j}{j!} = \exp{\left(\lambda m^{-1/2}\right)}$ Then $$\sum_{j\geq 0}\frac{(\lambda m^{-1/2})^j}{j!} \exp{\left(-B\,m^{-j}\right)}\leq \sum_{j\geq 0}\frac{(\lambda m^{-1/2})^j}{j!} = \exp{\left(\lambda m^{-1/2}\right)}\,.$$
Here's what I suggest you do to try and find a closed form, or perhaps a less desirable but still maybe useful a good approximation: Symbolic Regression. Various statistical toolboxes have algorithms for this but you can try making such a tool yourself. It is basically a method of building mathematical models to fit data by randomly generating functions of the inputs and comparing the outputs of these functions to the values of in your data. So in your case, the inputs would be $B$, $\lambda$, and $m$ and your output would be the value of the series at various combinations of your inputs. You'll need to approximate the sum at different inputs to generate your output data. Then feed this data into a symbolic regression algorithm and let it run for a long time. You'll get (probably) a good estimator if anything. I doubt there is a closed form. But if there is, this is likely the best way about going to find it. Here's a Wiki page on the topic: https://en.wikipedia.org/wiki/Symbolic_regression