# Riemann Zeta Function in terms of Dirichlet Eta Function

## 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 \ \$ $\ds \map \zeta s$ $=$ $\ds \frac 1 {1 - 2^{1 - s} } \map \eta s$

$\blacksquare$