Ordinary generating function for Bernoulli polynomial I know the exponential generating function for the Bernoulli polynomial $B_n(x)$:$$\frac{te^{tx}}{e^t-1}=\sum_{n=0}^\infty B_n(x)\frac{t^n}{n!}.$$
But is there an ordinary generating function? i.e a function $f$ such that $$f(t,x)=\sum_{n=0}^\infty B_n(x)t^n$$ valid for some open interval of $x$ and some open interval of $t$.
 A: No. The exponential generating function has a pole at $t = 2 \pi i$, so for any $x$, $B_n(x)$ grows at least as fast as about $\frac{n!}{(2\pi)^n}$, hence the ordinary generating function has zero radius of convergence in $t$. 
When possible, you can convert exponential generating functions to ordinary ones using a variant of the Laplace transform. 
A: I think you should actually be able to express this generating function in terms of polygamma functions. In particular, we notice that an ordinary (yes, very atypical) generating function for the Bernoulli numbers following from Borel summation (see here) is given in terms of $\psi^{(1)}(z)$ (the trigamma function) as 
$$\widetilde{B}(z) = \sum_{n \geq 0} B_n z^n = \sum_{k \geq 1} \frac{z}{(zk+1)^2} = \frac{1}{z} \psi_1(1/z)-z.$$
Now we can easily see from the binomial transform that 
$$B_n(x) = \sum_{k=0}^n \binom{n}{k} B_k x^{n-k} = x^n \cdot [z^n] \frac{1}{1-xz} \widetilde{B}\left(-\frac{z}{x(1-z)}\right) = [z^n] \frac{1}{1-z} \widetilde{B}\left(-\frac{z}{1-xz}\right).$$ 
So your desired ordinary generating function is expressed as 
$$\frac{1}{1-xz}\left[\frac{xz-1}{z} \psi_1\left(\frac{xz-1}{z}\right) + \frac{z}{1-xz}\right].$$
