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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?

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up vote 2 down vote accepted

I'd be inclined to construe "radius of convergence" as meaning the radius of convergence as a power series in $t$ rather than in $x$, since the $x$ is part of what it's a generating function of.

The numerator in $te^{tx}/(e^t-1)$, as a function of $t$ is an entire function, so the only singularities will be where the denominator is $0$, and then not necessarily if the numerator is also $0$ there. The numerator is $0$ at $t=0$. At that point the top and bottom both of a zero of multiplicity $1$, so its a removable singularity. So it is indeed a power series (you're right-hand side is right). The other places where $e^t-1$ is $0$ are $\pm n\pi$ for $n=1,2,3,\ldots$. The distance from $0$ to the nearest of those is $2\pi$, so that is the radius of convergence.

If you look at the left side as a function of $x$, it's an entire function, so as a power series in $x$ you get radius $\infty$.

What happens when you view it as a function of both $x$ and $t$ is more than I will attempt to say anything about here. (I haven't thought about that.)

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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 $2\pi$.

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