I am reading about general dirichlet series from my class notes and I also referred to wikipedia here:
https://en.m.wikipedia.org/wiki/General_Dirichlet_series
In the 2nd section : Abcissa of convergence I am unable to follow with this: If $\sigma_{c} \leq 0$ then $\sum a_n$ converges. But how in this case $\sigma_{c} $ is given by $\sigma_{c} = lim n\to \infty sup \frac{ log |a_{n+1} + a_{n+2}+... | } {\lambda_{n}} $, which is clearly different from the case in which $\sigma_{c} \geq 0$.
In this answer, we are going to prove
Proposition: Suppose the series $\sum_n a_ne^{-\lambda_ns}$ converges for some $s=s_0$ with $\sigma_0=\Re(s_0)<0$, then
$$ \limsup_{N\to\infty}{\log|a_N+a_{N+1}+\cdots|\over\lambda_N}\le\sigma_0 $$
Proof. For convenience, let's define
$$ A(x)=\sum_{n\le x}a_n $$
and
$$ S(x)=\sum_{n\le x}a_ne^{-\lambda_n s_0} $$
Then by the assumptions, we know that there exists $K>0$ such that $|S(x)|\le K$ holds for all positive $x$. Thus, if we let $N>M$ be two positive integers, then
$$ A(N)-A(M)=\sum_{M<n\le N}a_n=\sum_{M<n\le N}[S(n)-S(n-1)]e^{\lambda_n s_0} $$
Now, it follows from summation by parts that
$$ A(N)-A(M)=S(N)e^{\lambda_Ns_0}-S(M)e^{\lambda_Ms_0}+\sum_{M\le n<N}S(n)[e^{\lambda_{n+1}s_0}-e^{\lambda_ns_0}] $$
Now, by triangle's inequality, we have
$$ \begin{aligned} |A(N)-A(M)| &\le2Ke^{\lambda_M\sigma_0}+K\sum_{M\le n<N}|e^{\lambda_{n+1}s_0}-e^{\lambda_ns_0}| \cr &\le2Ke^{\lambda_M\sigma_0}+K|s_0|\int_{\lambda_M}^{\lambda_N}e^{\sigma_0u}\mathrm du \cr &\le K\left(2+\left|\frac{s_0}{\sigma_0}\right|\right)e^{\lambda_M\sigma_0} \end{aligned} $$
Since the resulting upper bound does not depend on $N$, we have that for each fixed $s_0$, there exists $B=B(s_0)>0$ such that
$$ |a_N+a_{N+1}+a_{N+2}+\cdots|\le Be^{\lambda_N\sigma_0} $$
Taking logarithms on both side concludes the proof.
It can be easily shown that the limit supremum is exactly the abscissa of convergence. Hope this can address your concern!