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

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Theorem

\(\ds \sum_{k \mathop = 1}^\infty \paren {\map \zeta {2k} - 1}\) \(=\) \(\ds \dfrac 3 4\)


Proof

Sum down each column, then sum across:


$\begin{array}{r|cccccccccc} \paren {\map \zeta {2k} - 1} & & & & & & & \\ \hline \paren {\map \zeta 2 - 1} & \paren {\dfrac 1 2}^2 & \paren {\dfrac 1 3}^2 & \paren {\dfrac 1 4}^2 & \cdots & \paren {\dfrac 1 n}^2 & \cdots \\ \paren {\map \zeta 4 - 1} & \paren {\dfrac 1 2}^4 & \paren {\dfrac 1 3}^4 & \paren {\dfrac 1 4}^4 & \cdots & \paren {\dfrac 1 n}^4 & \cdots \\ \paren {\map \zeta 6 - 1} & \paren {\dfrac 1 2}^6 & \paren {\dfrac 1 3}^6 & \paren {\dfrac 14}^6 & \cdots & \paren {\dfrac 1 n}^6 & \cdots \\ \paren {\map \zeta 8 - 1} & \paren {\dfrac 1 2}^8 & \paren {\dfrac 1 3}^8 & \paren {\dfrac 1 4}^8 & \cdots & \paren {\dfrac 1 n}^8 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \hline \ds \sum_{k \mathop = 1}^\infty \paren {\map \zeta {2k} - 1} & \paren {\dfrac 1 1} \paren {\dfrac 1 3} & \paren {\dfrac 1 2} \paren {\dfrac 1 4} & \paren {\dfrac 1 3} \paren {\dfrac 1 5} & \cdots & \paren {\dfrac 1 {n - 1} } \paren {\dfrac 1 {n+1} } & \cdots \end{array}$


First, summing down the array:

\(\ds \paren {\dfrac 1 {n^2} } + \paren {\dfrac 1 {n^2} }^2 + \paren {\dfrac 1 {n^2} }^3 + \cdots\) \(=\) \(\ds \paren {\dfrac 1 {n^2} } \paren {\dfrac 1 {1 - \dfrac 1 {n^2} } }\) Sum of Infinite Geometric Sequence: Corollary $1$
\(\ds \) \(=\) \(\ds \paren {\dfrac 1 {n^2} } \paren {\dfrac {n^2} {n^2 - 1} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 {n^2 - 1}\)


Next, summing across, we note:

\(\ds \dfrac 1 {n^2 - 1}\) \(=\) \(\ds \dfrac 1 2 \paren {\dfrac 1 {n - 1} - \dfrac 1 {n+1} }\)


Therefore,

\(\ds \paren {\dfrac 1 1 } \paren {\dfrac 1 3 } + \paren {\dfrac 1 2 } \paren {\dfrac 1 4 } + \paren {\dfrac 1 3 } \paren {\dfrac 1 5 } + \cdots\) \(=\) \(\ds \dfrac 1 2 \paren {\paren {\dfrac 1 1 - \dfrac 1 3 } + \paren {\dfrac 1 2 - \dfrac 1 4 } + \paren {\dfrac 1 3 - \dfrac 1 5 } + \cdots }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\dfrac 1 1 + \dfrac 1 2 - \paren {\dfrac 1 3 - \dfrac 1 3 } - \paren {\dfrac 1 4 - \dfrac 1 4 } - \cdots }\) regroup terms
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\dfrac 3 2 - \paren 0 - \paren 0 - \paren 0 - \cdots }\)
\(\ds \) \(=\) \(\ds \dfrac 3 4\)

Hence:

\(\ds \sum_{k \mathop = 1}^\infty \paren {\map \zeta {2k} - 1}\) \(=\) \(\ds \dfrac 3 4\)

$\blacksquare$


Also see


Sources

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