Let $f:\mathbb{N}\to\mathbb{C}$ be a function for which there exists a constant $0<A\in\mathbb{R}$ such that: $\lim_{x\to\infty}\frac{1}{x}\sum_{n\le x}f(n)=A.$
Show that: $\lim_{x\to\infty}\sum_{n\le x}f(n)\log(n)=Ax\log(x)(1+o(1)).$
I've been able to deduce the following:
1)For $0<M\in\mathbb{R}$, a function $g:[M,\infty)\to\mathbb{R}$ which is continuous on $[M,\infty)$ is said to be slowly oscillating if $\forall\epsilon>0\quad\lim_{x\to\infty}\frac{f(x\epsilon)}{f(x)}=1.$
Now, proposition 1.9 in De-Koninck&Luca states that a function $f$ that is once-continuously differentiable on $[M,\infty)$ is slowly oscillating iff $\lim_{x\to\infty}\int_M^x\frac{dt}{f(t)}=\frac{x}{f(x)}(1+o(1))$.
It's easy enough to verify from the definition that the function: $\frac{1}{A\log(x)}$ is slowly oscillating, and hence by proposition (1.9) $\quad\lim_{x\to\infty}\int_1^xA\log(t)dt=Ax\log(x)(1+o(1)).$ This can obviously be seen by a direct calculation but I wish to emphasize the apparent necessity of proposition (1.9) in the solution to my question.
2)$\frac{f(n)}{n}\to 0$. Maybe useful but I can't see how.
I'm unable to use the first statement of the question in order to deduce any conditions on $f(n)$ which might allow us to approximate it using an integral, or maybe translate between the appearances of $\frac{1}{x}$ and $\log(n)$, although these expressions always hang out together so it seems like a natural idea.
Working back from my first deduction has not led me too far either. A full solution, or a sequence of (not too vague please) hints will both be appreciated.
Thank you in advance.