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

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Theorem

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


Proof

Sum down each column, then sum across:


$\begin{array}{r|cccccccccc}

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

\paren {\map \zeta 4 - 1} & \paren {\dfrac 1 2}^4 & \paren {\dfrac 1 3}^4 & \paren {\dfrac 14}^4 & \cdots & \paren {\dfrac 1 n}^4 & \cdots \\

\paren {\map \zeta 5 - 1} & \paren {\dfrac 1 2}^5 & \paren {\dfrac 1 3}^5 & \paren {\dfrac 1 4}^5 & \cdots & \paren {\dfrac 1 n}^5 & \cdots \\

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


First, summing down the array:

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


Next, summing across, we note:

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


Therefore,

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

Hence:

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

$\blacksquare$


Also see


Sources

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