Riemann and Bernoulli join forces

We know for positive n $\zeta(2n)=\dfrac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}$ Hence we have $\sum_{n=1}^\infty \frac{\zeta(2n)}{B_{2n}}=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}(2\pi)^{2n}}{2(2n)!}\\=\dfrac{-1}{2}\left(\sum_{n=0}^\infty \left(\dfrac{(-1)^{n}(2\pi)^{2n}}{(2n)!}\right)-1\right)$ The summation is the taylor series for $\cos(x)=\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!}$ . So : $\dfrac{-1}{2}(\cos(2\pi)-1)=\dfrac{-1}{2}(1-1)=\boxed{0}$