Here $\chi$ is a completely multiplicative function with $\chi(p)\in\{-1,+1\}$ for any prime $p$, but not necessarily a character. Also $s=\sigma + it$ as usual. The series corresponding to $L(s,\chi)$ is the expansion of the Euler product $$\prod_{p\in P} \frac{1}{1-\chi(p)p^{-s}} =\sum_{k=1}^\infty \frac{a_k}{k^s}$$ where $P$ is an infinite subset of primes (or the full set), and $2\in P$. The product is ordered by increasing values of $p$. The $\eta(s,\chi)$ series is the alternating version: $$ \sum_{k=1}^\infty (-1)^{k+1}\frac{a_k}{k^s} $$ When $\sigma>1$ and everything converges nicely, we have: $\eta(s,\chi)=(1-2^{1-s})L(s,\chi)$. Note that $a_k\in\{-1,0,+1\}$.
The goal is to obtain a partial analytic continuation of one series (usually not to the full complex plane) when the other one (its alternating version) converges over a larger domain. The fundamental example is when $\chi(p)=1$. Then $L$ is the Riemann zeta function defined for $\sigma>1$ due to convergence limitations ($a_k=1$), while $\eta$ is the Dirichlet eta function defined (via the aternating series) for $\sigma > 0$.
The purpose is to work more easily with some $\chi$ that causes one of the two series to have a smaller domain of convergence than its alternating version. Or is it possible that both series have abscissa of convergence $\sigma=1$, depending on $\chi$?
