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

\(\ds \leadsto \ \ \)

\(\ds \paren {1 - 2^{1 - s} } \map \zeta s\)

\(=\)

\(\ds \map \eta s\)

\(\ds \leadsto \ \ \)

\(\ds \map \zeta s\)

\(=\)

\(\ds \frac 1 {1 - 2^{1 - s} } \map \eta s\)

$\blacksquare$

Also see

• Analytic Continuation of Riemann Zeta Function using Dirichlet Eta Function