# Relation between Zeta and Dirichlet eta functions proof

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

$$\textit{proof}$$

We will start by the RHS

$$\left( 1-2^{1-s} \right) \zeta(s) = \zeta(s) – 2^{1-s} \zeta(s)$$

Which can be written as sums of series

$$\sum_{n=1}^\infty \frac{1}{n^s} – \frac{1}{2^{s-1}}\sum_{n=1}^\infty \frac{1}{n^s}$$

$$\sum_{n=1}^\infty \frac{1}{n^s} – 2\sum_{n=1}^\infty \frac{1}{(2n)^s}$$

Clearly we can see that we are subtracting even terms twice , this is equivalent to

$$\sum_{n=1}^\infty \frac{1}{(2n-1)^s} – \sum_{n=1}^\infty \frac{1}{(2n)^s}$$

This looks easier to understand if we write the terms

$$\left(1+\frac{1}{3^s}+\frac{1}{5^s}+ \cdots\right) – \left( \frac{1}{2^s}+\frac{1}{4^s} + \frac{1}{6^s}+\cdots \right)$$

Rearranging the terms we establish the alternating form

$$1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} =\eta(s) \,$$

This entry was posted in Dirichlet eta function, Zeta function and tagged , , . Bookmark the permalink.