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 | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {1 - 2^{1 - s} } \map \zeta s\) | \(=\) | \(\ds \map \eta s\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \zeta s\) | \(=\) | \(\ds \frac 1 {1 - 2^{1 - s} } \map \eta s\) |
$\blacksquare$