Recursive formula for Bernoulli numbers from power series

Question

This is Ch 7, Exercise 62 in Palka's Complex Function Theory. Define $$ f:B(0,2 \pi) \to \mathbb{C},z \mapsto \frac{z}{e^z-1}\text{ if }z \neq 0, \; f(0):=1$$ Let $f$ have Taylor series $$f(z)= \sum_{n=0}^\infty \frac{B_n}{n!}z^n$$ where the $B_n$ are Bernoulli numbers (Here I use Wikipedia's notation, not Palka's). The question is to show $$\sum_{k=1}^{n}\binom{2n+1}{2k}B_{2k}= \frac{2n-1}{2}\tag{1}$$ I am stuck on showing this, but have some initial leads (from earlier in the problem). We have $B_0=f(0)=1$, and from Palka's hint, $z \mapsto z/2+f(z)$ is even: $$\begin{split} \frac{z}{2}+f(z)=& \frac{z(e^z-1)+2z}{2(e^z-1)}= \frac{ze^z+z}{2(e^z-1)}\\ \frac{-z}{2}+f(-z)=& \frac{-z}{2}- \frac{ze^z}{1-e^z}= \frac{-z(e^z-1)+2ze^z}{2(e^z-1)}= \frac{ze^z+z}{2(e^z-1)} \end{split}$$ which reveals $B_1=-1/2$ and $B_n=0$ for $n>2$ odd. I have also tried manipulating $(1)$: since $B_1$ is the only nonzero odd Bernoulli number: $$\sum_{k=1}^{2n}\binom{2n+1}{k}B_k= \frac{2n-1}{2}+ \binom{2n+1}{1}B_1=-1$$ or starting the sum at $0$: $$\sum_{k=0}^{2n}\binom{2n+1}{k}B_k=-1+ \binom{2n+1}{0}B_0=0$$ but I didn't get any additional insight on how to show $(1)$. Any help is greatly appreciated.

Answer 1

Hint:

$$(e^z-1)f(z)=z.$$

Differentiating $n$ times,

$$\sum_{k=0}^n\binom nk(e^z-1)^{(n-k)}f^{(k)}(z)=0.$$

Then with $z=0$,

$$\sum_{k=0}^n\binom nkF_k-F_n=0.$$

Answer 2

Here is a sketch, where I will omit details for you to flesh out. Recall that $\dbinom{n}{k}=\dbinom{n}{n-k}$, and rewrite the sum under consideration as

$$\begin{align*} \sum_{k=1}^{n}\binom{2n+1}{2k}B_{2k}&=\sum_{k=1}^{n}\binom{2n+1}{2n-2k+1}B_{2k}\\ &=\sum_{k=1}^{n}\frac{(2n+1)!}{(2n-2k+1)!}\frac{B_{2k}}{(2k)!}\\ &=-1+\sum_{k=0}^{n}\frac{(2n+1)!}{(2n-2k+1)!}\frac{B_{2k}}{(2k)!} \end{align*}$$

where the last sum should be recognizable as a convolution. Recall that multiplying two power series yields a new power series whose coefficients are a convolution of the coefficients of the two original series, and proceed from there.