Note to the behavior of the maximal term of Dirichlet series absolutely convergent in half-plane

By $S_0(\Lambda)$ denote a class of Dirichlet series $F(s)=\sum_{n=0}^{\infty}a_n\exp\{s\lambda_n\} (s=\sigma+it)$ withan increasing to $+\infty$ sequence $\Lambda=(\lambda_n)$ of exponents ($\lambda_0=0$) and the abscissa of absolute convergence $\sigma_a=0$.We say that $F\in S_0^*(\Lambda)$ if $F\in S_0(\Lambda)$ and $\ln \lambda_n=o(\ln |a_n|)$ $(n\to\infty)$. Let$\mu(\sigma,F)=\max\{|a_n|\exp{(\sigma\lambda_n)}\colon n\ge 0\}$ be the maximal term of Dirichlet series. It is proved that in order that $\ln (1/|\sigma|)=o(\ln \mu(\sigma))$ $(\sigma\uparrow 0)$ for every function $F\in S_0^*(\Lambda)$ it is necessary and sufficient that $\displaystyle \varlimsup\limits_{n\to\infty}\frac{\ln \lambda_{n+1}}{\ln \lambda_n}<+\infty. $For an analytic in the disk $\{z\colon |z|<1\}$ function $f(z)=\sum_{n=0}^{\infty}a_n z^n$ and $r\in (0, 1)$ we put $M_f(r)=\max\{|f(z)|\colon |z|=r<1\}$ and $\mu_f(r)=\max\{|a_n|r^n\colon n\ge 0\}$. Then from hence we get the following statement: {\sl if there exists a sequence $(n_j)$ such that $\ln n_{j+1}=O(\ln n_{j})$ and $\ln n_{j}=o(\ln |a_{n_{j}}|)$ as $j\to\infty$, then the functions $\ln \mu_f(r)$ and $\ln M_f(r)$ are or not are slowly increasing simultaneously.