Prove that the analytic mean value of an arithmetic function equals the logarithmic mean value Let f be an arithmetic function and let $F(s)$ be its Dirichlet series $\sum_{n=1}^{\infty}f(n)n^{-s}$. We say f has an analytic mean value A if
$F(s)=\frac{A}{s-1}+o(\frac{1}{s-1})$
as $s\rightarrow 1^+$. My question is, how would we show that the existence of the logarithmic mean value for f implies the existence of the analytic mean value, i.e. if the following limit exists:
$\lim{x\to\infty}\frac{1}{logx}\sum_{n\leq x}\frac{f(n)}{n}$
then the analytic mean value exists, and the two values are equal? So far I have tried using summation by parts to get the following:
$\sum_{n\leq x}f(n)n^{-s}=x^{1-s}\sum_{n\leq x}\frac{f(n)}{n}+\int_{1}^{x}\frac{(s-1)\sum_{n\leq t}\frac{f(n)}{n}}{t^{s-2}}dt$
This appears to get us part of the way there, since there is an $f(n)/n$ sum on the right hand side which looks similar to the logarithmic mean value (minus the log), but I'm not sure where to go from there. Any help is appreciated!

For convenience, let
$$
B(x)=\sum_{n\le x}{f(n)\over n}=A\log x+o(\log x).
$$
Then for every $u>0$ there is
$$
F_T(1+u)=\sum_{n\le T}{f(n)\over n^{1+u}}=\int_{1^-}^T{\mathrm dB(t)\over t^u}={B(T)\over T^u}+u\int_1^T{B(t)\over t^{u+2}}\mathrm du.
$$
Since $B(T)=O(T)=o(T^u)$, we see that for all $u>0$, as $T\to+\infty$ there is
$$
F(1+u)=u\int_1^\infty{B(t)\over t^{u+1}}\mathrm dt=Au\underbrace{\int_1^\infty{\log t\over t^{u+1}}\mathrm dt}_{I_1}+u\underbrace{\int_1^\infty{B(t)-A\log t\over t^{u+1}}\mathrm dt}_{I_2}
$$
For $I_1$, substitution gives
$$
I_1=\int_0^\infty ye^{-uy}\mathrm dy={1\over u^2}.
$$
For $I_2$, by the conditions for $B(x)$ we know that
$$
\forall\varepsilon>0,\exists X>0,(t>X\Rightarrow|B(t)-A\log t|<\frac12\varepsilon\log t)
$$
and
$$
\exists M>0,(1\le t\le X\Rightarrow|B(t)-A\log t|\le M\log t),
$$
so we see that $I_2$ is bounded by
\begin{aligned}
|I_2|
&<Mu\int_1^X{\log t\over t^{u+1}}\mathrm du+\frac12\varepsilon u\int_X^\infty{\log t\over t^{u+1}}\mathrm du \\
&\le Mu\int_1^X{\log t\over t}\mathrm dt+\frac12\varepsilon u\int_1^\infty{\log t\over t^{u+1}} \\
&=\frac12 Mu(\log X)^2+{\varepsilon\over2u}.
\end{aligned}
Now, let $0<u<\varepsilon/(M(\log X)^2)$, so we have $|I_2|<\varepsilon/u$. Combining all these results, we see that
$$
F(s)=\sum_{n\ge1}{f(n)\over n^s}={A\over s-1}+o\left(1\over|s-1|\right).
$$
