The Dirichlet eta series is holomorphic on $\mathrm{Re}s>0$ and equals the prefactor times zeta there

Choose $\sigma >0$ and $M\ge 0$ so that $\mathrm{Re}s\ge \sigma$ and $\mathrm{\mid }s\mathrm{\mid }\le M$ on $K$. Pair the alternating series as $\eta (s)={\sum }_{n=1}^{\mathrm{\infty }}((2n-1{)}^{-s}-(2n{)}^{-s})\mathrm{.}$ Using [A1] and [L4], $\mathrm{\mid }(2n-1{)}^{-s}-(2n{)}^{-s}\mathrm{\mid }=\mid s{\int }_{2n-1}^{2n}{x}^{-s-1}dx\mid \le M(2n-1{)}^{-\sigma -1}\mathrm{.}$ Taking $\sigma$ rational, [L5] makes ${\sum }_n(2n-1{)}^{-\sigma -1}$ convergent. Therefore [L3] gives local uniform convergence of the paired series on $K$, so $\eta$ is holomorphic on $\mathrm{Re}s>0$.

On $\mathrm{Re}s>1$, the zeta series of [L1] converges absolutely, so regrouping odd and even terms gives $\eta (s)={\sum }_{n\ge 1}{n}^{-s}-2{\sum }_{n\ge 1}(2n{)}^{-s}=(1-2^{1-s})\zeta (s)\mathrm{.}$

By step 1.1, $\eta$ is holomorphic on $\mathrm{Re}s>0$. By [L2], the function $(1-2^{1-s})\zeta (s)$ is also holomorphic there: the factor $1-2^{1-s}$ vanishes at $s=1$ and removes the only pole of $\zeta$. Step 1.2 shows that the two holomorphic functions agree on the nonempty open set $\mathrm{Re}s>1$, so [A2] gives the identity $\eta (s)=(1-2^{1-s})\zeta (s)$ throughout $\mathrm{Re}s>0$.