sequences and series - Why does $sum_{k=1}^infty (ζ[2k+1]-1)=frac{1}{4}$

Can someone explain why

$$\sum_{k=1}^\infty (ζ[2k+1]-1)=\frac{1}{4}?$$

Answer

We can rearrange
$$\begin{align}\sum_{k=1}^\infty(\zeta(2k+1)-1)&=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac1{n^{2k+1}}\\ &=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{n^{2k+1}}\\ &=\sum_{n=2}^\infty\frac1{n(n^2-1)}\\ &=\sum_{n=2}^\infty\left(\frac1{2n(n-1)}-\frac1{2(n+1)n}\right)\\&=\frac14\end{align}$$