Is the series $‎\sum_{n=0}^{‎\infty‎}‎\frac{B_n(z)}{2^n}‎$ convergent? ‎$B_n(z)‎‎$‎‎ ‎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?
 A: 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 $0$.
