I am trying to get an asymptotic formula such as
$$ L_4(s, n) \sim L_4(s) + \rho_n(s)\Lambda_n + \frac{\alpha(s)}{\sqrt{n}} + \frac{\beta(s)}{\sqrt{n\log n}}+\cdots$$
where $L_4(s, n)$ is the first $n$ factors of the Euler product for the $L_4(s)$ Dirichlet function with non-trivial character $\chi_4$. There are a few potential candidates for $\Lambda_n$. Here $s=\sigma+it$, with $t=0$ and roughly speaking, $0.80 < \sigma < 1.10$ (this is where the approximation works best, especially around $\sigma = 0.90$). Once $\Lambda_n$ is fixed, the choice for $\rho_n(s)$ is obvious and it sounds like $\rho_n(s)\rightarrow \rho(s)$, a constant depending only on $s$ as $n\rightarrow\infty$.
This works if $\Lambda_n \rightarrow 0$. And this seems to be the case in all the examples tested. Very few choices work well for $\Lambda_n$, but the one that works best so far is this: $$\Lambda_n = \frac{1}{f(n)}\sum_{k=1}^n \chi_4(p_k),$$
where $p_1, p_2, p_3$ are the standard primes with $p_1=2$. Convergence status is known if $f(n) = n\log n$, but then the approximation is not great. The best so far is $f(n)=n$ which leads to smooth behavior in the specified interval for $\sigma$. With this choice, do we have convergence for $\Lambda_n$, and in addition, do we have $\Lambda_n \rightarrow 0$?
If you wonder what $\rho_n(s)$ is, it is the ratio $\tau_n(s)/\nu_n(s)$ where the numerator and denominator are the standard deviations, computed on the first $n$ values respectively of $L_4(s, n)$ and $\Lambda(n)$. This choice is obvious and aims at optimizing the fit between $L_4(s,n)-L_4(s)$, and $\Lambda_n$.
