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:
$$\Lambda_n = \sum_{k=1}^n \frac{\chi_4(p_k)}{f(k)},$$

where $p_1, p_2, p_3$ are the standard primes with $p_1=2$. Convergence status is known if $f(k) =p_k$, but then the approximation is not great (it's a bit chaotic). It gets better with $f(k) = p_k^s$ but then $\Lambda_n$ also depends on $s$.
The best so far is $f(k)=k$ 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 $\Lambda_n$.