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 $\zeta(s) = \dfrac 1 {1 - 2^{1-s}} \eta \left({s}\right)$.


Proof

\(\displaystyle \zeta\left(s\right) - \eta\left(s\right)\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^s} - \sum_{n\mathop = 1}^\infty \frac{(-1)^{n-1} }{n^s}\) $\quad$ Definition of Riemann Zeta Function, Definition of Dirichlet Eta Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left( \frac 1 {n^s} +\frac{(-1)^n}{n^s} \right)\) $\quad$ Linearity of Series $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2\sum_{n=1}^\infty \frac 1 {\left(2n\right)^s}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2^{1-s}\sum_{n=1}^\infty \frac 1 {n^s}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2^{1-s}\zeta\left(s\right)\) $\quad$ Definition of Riemann Zeta Function $\quad$

Rearranging,

\(\displaystyle \left(2^{1-s}-1\right)\zeta\left(s\right)\) \(=\) \(\displaystyle -\eta\left(s\right)\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \zeta\left(s\right)\) \(=\) \(\displaystyle \frac 1 {1-2^{1-s} } \eta\left(s\right)\) $\quad$ $\quad$

$\blacksquare$

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