$B_n(z)$ is the Bernoulli polynomial, we know that the series
$$\sum_{n=0}^{\infty}\frac{B_n(z)}{n!}$$
is convergent for every $z\in\mathbb{C}$ and its closed form is equal to $$\frac{e^z}{e-1}.$$ Now I deal to the series $$\sum_{n=0}^{\infty}\frac{B_n(z)}{2^n}$$ in my research.
(a) Is the above series convergent? If yes, for which $z$?
(b) Does the series have any closed form?
(c) Does there exist any method or related books or similar reference?
Formally, the o.g.f. comes from the Laplace transform of the e.g.f. That is, given $$ \frac{t e^{zt}}{e^t-1} = \sum_{n=0}^\infty B_n(z) \frac{t^n}{n!}$$ $$ \sum_{n=0}^\infty B_n(z) s^{-1-n} = \int_0^\infty \frac{t e^{zt}}{e^t-1} e^{-st}\; dt = \Psi^{(1)}(1-z+s) $$ The right side is singular when $1-z+s$ is a nonpositive integer. If the series had a positive radius of convergence, the set of singularities would have to be bounded. Since it is not, we conclude that the radius of convergence is $0$. In particular, your series $\sum_{n=0}^\infty B_n(z) 2^{-n}$ diverges for all $z$.
EDIT: There' a simpler argument. The fact that the e.g.f. $\frac{t e^{zt}}{e^t-1}$ has finite radius of convergence (as it has singularities $t = 2 n \pi i$ for nonzero integers $n$) means that $\limsup_n |B_n| r^n/n! > 0$ for some $r$. But that says for some $r$ and $\epsilon > 0$, $|B_n| \ge \epsilon n! r^{-n}$ infinitely often. For any $z \ne 0$, $\epsilon n! |z/r|^n \to \infty$, so your series can't converge.
More generally, whenever the e.g.f. of a sequence has a finite radius of convergence, the o.g.f. has radius of convergence $0$.