Sum over k from 1 to Infinity of Zeta of 2k Minus One

Proof
Sum down each column, then sum across:


 * $\begin{array}{r|cccccccccc}

\paren {\map \zeta {2k} - 1} & &  &  &  &  &  &  \\ \hline

\paren {\map \zeta 2 - 1} & \paren {\dfrac 1 2}^2 & \paren {\dfrac 1 3}^2 & \paren {\dfrac 1 4}^2 & \cdots & \paren {\dfrac 1 n}^2 & \cdots \\

\paren {\map \zeta 4 - 1} & \paren {\dfrac 1 2}^4 & \paren {\dfrac 1 3}^4 & \paren {\dfrac 1 4}^4 & \cdots & \paren {\dfrac 1 n}^4 & \cdots \\

\paren {\map \zeta 6 - 1} & \paren {\dfrac 1 2}^6 & \paren {\dfrac 1 3}^6 & \paren {\dfrac 14}^6 & \cdots & \paren {\dfrac 1 n}^6 & \cdots \\

\paren {\map \zeta 8 - 1} & \paren {\dfrac 1 2}^8 & \paren {\dfrac 1 3}^8 & \paren {\dfrac 1 4}^8 & \cdots & \paren {\dfrac 1 n}^8 & \cdots \\

\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \hline \sum_{k \mathop = 1}^\infty \paren {\map \zeta {2k} - 1} & \paren {\dfrac 1 1} \paren {\dfrac 1 3} & \paren {\dfrac 1 2} \paren {\dfrac 1 4} & \paren {\dfrac 1 3} \paren {\dfrac 1 5} & \cdots & \paren {\dfrac 1 {n - 1} } \paren {\dfrac 1 {n+1} } & \cdots \end{array}$

First, summing down the array:

Next, summing across, we note:

Therefore,

Hence: