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Define the Bernoulli numbers $B_n$ by $\frac{z}{2} \cot (z/2) = 1 - B_1 \frac{z^2}{2!} - B_2 \frac{z^4}{4!} - B_{3} \frac{z^6}{6!} - ...$ Explain why there are no odd terms in this series. What is the radius of convergence of the series? Find the first five Bernoulli numbers.

I understand why the power series expansion of $\frac{z}{2} \cot (\frac{z}{2})$ contains only terms with even powers (it has to do with the fact that the function is even, and so all the derivatives of odd order will vanish at $0$, so they odd order terms will vanish too). The only two things tricking are the radius of convergence and find the Bernoulli numbers.

First, let me discuss the radius of convergence. The form of the power series suggests that it's an expansion of $\frac{z}{2} \cot (\frac{z}{2})$ at $0$, but what's confusing me is that $\frac{z}{2} \cot (\frac{z}{2})$ is not defined at $0$. In fact, this leads to my confusion about the radius of convergence. From my understanding, the radius of convergence is the distance from the expansion point to the nearest pole, but $0$ is the nearest pole so shouldn't the radius of convergence be $0$?

As for calculating the Bernoulli numbers, my book discusses how to calculate the first few terms of the power series of $\frac{1}{g(z)}$ when $g(z)$ is analytic at $z=0$. I thought of using this calculation, but my problem is that $\frac{z}{2} \cot (\frac{z}{2})$ is not even defined at $z=0$. Also, in my book they calculate the power series of $\tan (z)$ (expanded at $z=0$ presumably). I thought of using the fact that $\cot (z) = \frac{1}{\tan (z)}$, but again the function is not defined at $z=0$. How am I to calculate the first five Bernoulli numbers?

J. W. Tanner
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user193319
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    There is a removable singularity at $z=0$ since $\lim_{z\to 0}z/\sin(z)=1$. You're correct about the radius; do you see now why it must be $2\pi$? – Integrand Oct 19 '20 at 15:16
  • @Integrand Any thoughts on how to calculate the Bernoulli numbers? – user193319 Oct 20 '20 at 01:17
  • Does this answer your question? $z \cdot \cot(z)$ series – Integrand Oct 20 '20 at 01:21
  • @Integrand This might help. But I just realized something: isn't the $n$-th coefficient of the power series expansion of $\frac{z}{2} \cot (\frac{z}{2})$ just given by taking the $n$-th derivative, evaluating at $z=0$, and dividing by $n!$? – user193319 Oct 20 '20 at 10:05
  • This answer maybe helpful https://math.stackexchange.com/questions/642625/calculating-bernoulli-numbers-from-sum-limits-n-0-infty-fracb-nxnn-f/642689#642689 – user91500 Oct 21 '20 at 09:28
  • Having the Bernoulli numbers defined as the coefficients of the power series for $\frac{z}{2}\cot\frac{z}{2}$ (and nothing more), it looks natural to compute the latter as a quotient $f(z)/g(z)$ of two known power series: for $f(z)=\cos\frac{z}{2}$ and $g(z)=\frac{2}{z}\sin\frac{z}{2}$. – metamorphy Oct 22 '20 at 06:33

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