Bernoulli numbers identity with binomial coefficient The generating function for the Bernoulli numbers $B_k$ is given by $f(z) = \frac{z}{e^z -1}= \sum_{k=0}^{\infty} \frac{B_k}{k!} z^k$. Applying the identity
$$1 = \frac{e^z -1}{z} \cdot \sum_{k=0}^{\infty} \frac{B_k}{k!} z^k$$ yields $\sum_{n=0}^{k} \binom{k+1}{n} B_n= 0$. Recall that we also have that $f(z)= 1- \frac{z}{2}+ \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}$ which implies that $$\frac{e^z -1}{z} \cdot \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}= 1- \frac{e^z -1}{z}(1- \frac{z}{2}).$$ I believe the indices got me all confused, so I keep getting stuck on how to show that this above equality gives us the identity $\sum_{n=0}^{k} \binom{2k+1}{2n} B_{2n}= \frac{1}{2} (2k+1)$ after expanding both sides of the equality. How should I proceed?

If we don't have to use the equation
$$\frac{e^z -1}{z} \cdot \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} z^{2k}= 1- \frac{e^z -1}{z}(1- \frac{z}{2}),$$
then we can simply use the already proven
$$\sum_{n=0}^k \binom{k+1}{n}B_n = 0$$
for $k > 0$ and insert $k = 2m$ to obtain
$$\sum_{n=0}^m \binom{2m+1}{2n}B_{2n} = - \sum_{n=0}^{m-1} \binom{2m+1}{2n+1}B_{2n+1},$$
which, since $B_1 = -\frac{1}{2}$ and $B_{2n+1} = 0$ for $n \geqslant 1$ simplifies to
$$\sum_{n=0}^m \binom{2m+1}{2n}B_{2n} = -B_1\binom{2m+1}{1} = \frac{1}{2}(2m+1).$$
