For any complex $x$ we define the funcitons $B_n(x)$ by the equation $$ \frac{ze^{xz}}{e^z-1} = \sum_{n=0}^\infty \frac{B_n(x)}{n!} z^n , \text{ where } |z|< 2 \pi . $$ Page 264, Apostol , Introduction to Analytic Number Theory.
The function on the LHS can be extended analytically to all of $\mathbb{C}$ for any fixed $x$. So doesn't the power series representation holds for all $z \in \mathbb{C}$? Why is there "$|z|< 2\pi $"?
The left-hand side has poles when $e^z=1$ (excluding the one at $z=0$, which is cancelled by the numerator). These are $z=2k\pi i, k \in \mathbb{Z}\setminus \{ 0 \}$. The closest of these to the origin are $2\pi i$, and since a power series converges in the disk with radius the distance to the nearest singularity, the series only converges for $\lvert z \rvert <2\pi$.