Radius of convergence of the Bernoulli polynomial generating function power series. The generating function of the Bernoulli Polynomials is: $$\frac{te^{xt}}{e^t-1}=\sum_{k=0}^\infty B_k(x)\frac{t^k}{k!}.$$
Would it be right to say that the radius of convergence of this power series is $2\pi$ ? I'm not sure since the power series above is in fact a double series: $$\sum_{k=0}^\infty\left(\sum_{j=0}^k {k\choose j}B_{k-j} x^j\right)\frac{t^k}{k!}.$$
What if I were to choose a fixed value for $x$? Would the radius be $2\pi$ then, even for the double power series?

For every fixed $x=c$, the radius of convergence of the power series is $2\pi$. This is because $$\frac{ze^{cz}}{e^z-1}$$ is analytic everywhere except at $z=i2\pi n, n=\pm 1,\pm 2,\cdots$ (not at $0$ though.) The disk $B(0,2\pi)$ is the smallest one centered at $0$ that contains a singularity on its boundary, so the radius of convergence is $2\pi$.
