Sum of General Harmonic Numbers in terms of Riemann Zeta Function/Lemma

Lemma for Sum of General Harmonic Numbers in terms of Riemann Zeta Function

$\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + x}^r} = \sum_{k \mathop = 0}^{n - 1} \sum_{j \mathop = 1}^\infty \frac 1 {\paren {n j - k + x }^r}$

where:

$r$ and $x$ are complex numbers

$n \in \Z_{>0}$.

Proof

\(\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + x}^r}\)

\(=\)

\(\ds \paren {\paren {\frac 1 {\paren {1 + x}^r} + \frac 1 {\paren {2 + x}^r} + \cdots + \frac 1 {\paren {\paren {n - 1} + x}^r} + \frac 1 {\paren {n + x}^r} } + \paren {\frac 1 {\paren {\paren {n + 1} + x}^r} + \cdots + \frac 1 {\paren {2n + x}^r} } + \cdots }\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {\frac 1 {\paren {n + x}^r} + \frac 1 {\paren {\paren {n - 1} + x}^r} + \cdots + \frac 1 {\paren {2 + x}^r} + \frac 1 {\paren {1 + x}^r} } + \paren {\frac 1 {\paren {2n + x}^r} + \cdots + \frac 1 {\paren {2 n - \paren {n - 1} + x}^r} } + \cdots }\)

reversing the order

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^{n - 1} \sum_{j \mathop = 1}^\infty \frac 1 {\paren {n j - k + x }^r}\)

$\blacksquare$