Riemann Zeta Function in terms of Dirichlet Eta Function

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Theorem

Let $\zeta$ be the Riemann zeta function.

Let $\eta$ be the Dirichlet eta function.

Let $s \in \C$ be a complex number with real part $\sigma > 1$.


Then $\map \zeta s = \dfrac 1 {1 - 2^{1 - s} } \map \eta s$.


Proof

\(\ds \map \zeta s - \map \eta s\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^s} - \sum_{n \mathop = 1}^\infty \frac{\paren {-1}^{n - 1} } {n^s}\) Definition of Riemann Zeta Function, Definition of Dirichlet Eta Function
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 {n^s} + \frac {\paren {-1}^n} {n^s} }\) Sum of Summations equals Summation of Sum
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^s}\)
\(\ds \) \(=\) \(\ds 2^{1 - s} \sum_{n \mathop = 1}^\infty \frac 1 {n^s}\)
\(\ds \) \(=\) \(\ds 2^{1 - s} \map \zeta s\) Definition of Riemann Zeta Function
\(\ds \leadsto \ \ \) \(\ds \paren {1 - 2^{1 - s} } \map \zeta s\) \(=\) \(\ds \map \eta s\)
\(\ds \leadsto \ \ \) \(\ds \map \zeta s\) \(=\) \(\ds \frac 1 {1-2^{1 - s} } \map \eta s\)

$\blacksquare$


Also see