I was just reading this question and answer: How will this equation imply PNT and it raised a whole new question:

Given that $\sum_{n\le x} \Lambda(n)=x+o(x)$, prove that $$\sum_{n\le x} \frac{\Lambda(n)}{n}=\log x-\gamma +o(1),$$ where gamma is the Euler constant.

The above question is for the bounty. The full version above has been bothering me. Gerry Myerson pointed out we can prove $\sum_{n\leq x } \frac{\Lambda(n)}{n} =\log x +O(1)$ from only chebyshev estimate. But this is not my question.

**My Attempts:** I tried partial summation I end up with something like $\sum_{n} \frac{\psi(n)-n}{n^2}$. for that to converge $o(n)$ isn't strong enough. To do it my way, I would need to assume $$\psi(x)-x = O\left(\frac{x}{(\log x)^{1+\epsilon}}\right)$$ so that it converges. (Otherwise it could be as big as $\log x$ which is no good)

Can we prove the above estimate using only the basic prime number theorem $\psi(x)-x=o(x)$? Why or why not? Thank you!

Please note that my question is about a subtlety with converge, and blunt partial summation doesn't seem to work.