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!
 A: 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
$$
F(s)=\sum_{n\ge1}{f(n)\over n^s}={A\over s-1}+o\left(1\over|s-1|\right).
$$
