Mean value and Logarithmic value for specific arithmetic function Define the Mean value for function $f$ as $\lim \limits_{x \to \infty} \frac{1}{x} \sum \limits_{n \leq x} f(n)$ if the limit exists denoted as $M_f$
Define the Logarithmic value for function $f$ as $\lim \limits_{x \to \infty} \frac{1}{\ln x} \sum \limits_{n \leq x} \frac{f(n)}{n}$ if the limit exists denoted as $L_f$
Its easy to prove that if $M_f$ exist then so does $L_f$ and they are equal, but the other way around is not true in general.
Let $F(s) = \sum \limits_{n=1}^{\infty} f(n) n^{-s}$ for all $s>1$, and we are given that in addition $F(s)$ satisfy $F(s) = \frac{A}{s-1} +o(\frac{1}{s-1})$ for $s \to 1^{+}$ and $A$ is a constant.
Using this prove that $M_f,L_f$ exists for $f$ and that $M_f =L_f=A$.
I tried the usual method like (Abel,Euler Summation) but in every case i end up with two limits , one for $x \to \infty$ and one for $s \to 1^{+}$ which makes the problem way harder than usual, i wish for simple method how to tacle this problem or reduces the two limits to one or none. 
 A: A compelling reason why you didn't manage to prove the assertion is that it isn't true.
Consider $f(n) = (-1)^{n+1}\cdot n$. Then
$$\sum_{n = 1}^m f(n) = (-1)^{m+1}\biggl\lfloor \frac{m+1}{2}\biggr\rfloor\,,$$
so $M_f$ doesn't exist. But $F(s) = \eta(s-1)$ is an entire function, hence
$$F(s) = \frac{0}{s-1} + o\biggl(\frac{1}{s-1}\biggr)$$
for $s \to 1$. This $f$ does however have a logarithmic mean value, $L_f = 0$. To get an example where even the logarithmic mean value doesn't exist, consider $g(n) = f(n)\log n$. We have
\begin{align}
\sum_{n = 1}^{2m} \frac{g(n)}{n} &= \sum_{n = 1}^{2m} (-1)^{n+1}\log n \\
&= \sum_{n = 1}^{2m} \log n - 2\sum_{k = 1}^m \log (2k) \\
&= \sum_{n = 1}^{2m} \log n - 2m\log 2 - 2\sum_{k = 1}^m \log (k) \\
&= 2m\log (2m) - 2m + \frac{1}{2}\log (2m) + \log \sqrt{2\pi} + O(m^{-1}) \\
&\quad - 2m\log 2 - 2m\log m + 2m - \log m - 2\log \sqrt{2\pi} + O(m^{-1}) \\
&= -\frac{1}{2}\log (2m) + \log \sqrt{\frac{2}{\pi}} + O(m^{-1})
\end{align}
and consequently
$$\sum_{n = 1}^{2m+1} \frac{g(n)}{n} = \frac{1}{2}\log (2m+1) + \log \sqrt{\frac{2}{\pi}} + O(m^{-1})\,.$$
Thus $L_g$ doesn't exist (and a fortiori $M_g$ doesn't exist), yet $G(s) = - \eta'(s-1)$ is entire, whence
$$G(s) = \frac{0}{s-1} + o\biggl(\frac{1}{s-1}\biggr)$$
for $s \to 1$.
The existence of an "analytic mean value", i.e. the relation
$$F(s) = \frac{A}{s-1} + o\biggl(\frac{1}{s-1}\biggr) \tag{$\ast$}$$
for $s \to 1^+$ is a strictly weaker condition than the existence of a logarithmic mean value (and that in turn is strictly weaker than the existence of an ordinary or arithmetic mean value).
What is true — and not too difficult to prove — is that $L_f = A$ implies $(\ast)$.
