I have to prove that $\forall k \geq 1$ $$ \sum_{n\leq x} \frac{f(n)}{n} = \frac{1}{(k+1)!} \log^{k+1} x + O(\log^k x), $$ where $$ \sum_{n\leq x} f (n) = \frac{x}{k!} \log^k x + O(x\, \log^{k-1}x). $$ I started using the partial summation formula (integral version) that says $$ \sum_{n=1}^N f(n)g(n) = F(N)g(N) - \int_1^N F(x)g^\prime(x)dx, $$ where $F(x) = \sum_{n\leq x} f(n)$.
I chose $g(x) = 1/x$ and $f(x) = f(x)$. So we have $$ \sum_{n\leq x} \frac{f(n)}{n} = \frac{1}{x} \sum_{n\leq x} f(n) + \int_1^x \sum_{n\leq y} f(n) \frac{1}{y^2} dy. $$ The first term is equal to $$ \frac{1}{k!} \log^k x + O(\log^{k-1}x), $$ while the integral is equal to $$ \int_1^x \frac{1}{k!} \frac{1}{y} \log^k y + O\left(\frac{1}{y}\log^{k-1}y\right) dy. $$ Now I don't know how to continue. I think the idea is correct. Someone can help me with this integral?
You want to show that it's true to first order in $\log^k(x)$. So just use l'Hopital's rule on $\lim_{x\rightarrow\infty}\frac{\log^{k+1}x}{\int_1^x(1/y)\log^k(y)dy}$, and you can also use this fact (after some extra additions) to deduce that the error is $O(\log^k(x))$.