Let $f$ be arithmetic function such that $f=1 *g$ ( Dirichlet convolution ), mean value of $f$ is denoted $M_f$ and is equal to the limit $\lim \limits_{x \to \infty} \frac{1}{x} \sum \limits_{n \leq x} f(n)$ if it exists.
If we assume that $\sum \limits_{n=1}^{\infty} \frac{|g(n)|}{n}$ converges then its easy to prove that $M_f = \sum \limits_{n=1}^{\infty} \frac{g(n)}{n}$.
Assume that $\sum \limits_{n=1}^{\infty} \frac{g(n)}{n}$ converges conditionally and that $\frac{1}{x}\sum \limits_{n \leq x} |g(n)| \leq L$ bounded, i have 2 questions :
1) Prove that under these conditions we have that $M_f = \sum \limits_{n=1}^{\infty} \frac{g(n)}{n}$ ?
2) There is a function $g$ such that $\sum \limits_{n=1}^{\infty} \frac{g(n)}{n}$ converges conditionally and $\frac{1}{x}\sum \limits_{n \leq x} |g(n)|$ is not bounded and $M_f$ does not exists ?
Any help with one or more of the questions is much appricated.
My attemp : for the second question i have no idea, for the first $\frac{1}{x} \sum \limits_{n\leq x} f(n) = \frac{1}{x} \sum \limits_{d \leq x} g(d) \lfloor \frac{x}{d} \rfloor$ using Dirichlet convolution and is equal to $ \sum \limits_{d \leq x} g(d)+O(\frac{1}{x})\sum \limits_{d\leq x}g (d)$ ,the first sum tends to $\sum \limits_{d=1}^{\infty} g(d)$ when $x \to \infty$ which is what i need to prove that is equal to $M_f$ so what remains is to prove that the second sum tends to zero.