Asymptotic behavior of $\sum_{n>x} \frac{\log n}{n^2}$ There is a well-known question that seeks the asymptotic behaviour of this function, for $x\geq 2$:
$$\sum_{n\leq x} \frac{\phi(n)}{n^2}.$$
See, for example, Apostol "Introduction to Analytic Number Theory" question 6 on page 71.
I have found three posts on stackexchange related to this question, namely post A,
post B, and
post C. Using methods available to someone who has just read Chapter 3 of Apostol's text, the best solution to this question would seem to be apatch's response to post A.
The method boils down to showing the following identity
$$\sum_{d>x}\frac{\mu(d)\log d}{d^2} = O\left(\frac{\log x}{x}\right).$$
This is where I get unstuck. The first step is clearly
$$\left|\sum_{d>x}\frac{\mu(d)\log d}{d^2}\right| \leq \sum_{d>x}\frac{\log d}{d^2}$$
but then what do you do with what remains? 
One solution I found simply jumps from here to $O(\frac{\log x}{x})$, which seems non-trivial. Apatch's solution in post A does the following:
$$\sum_{d>x}\frac{\log d}{d^2}=\sum_{d>x}\frac{\log d}{d^\frac{1}{2}}.\frac{1}{d^\frac{3}{2}}<\frac{\log x}{x^\frac{1}{2}}\sum_{d>x}\frac{1}{d^\frac{3}{2}}$$
but I don't see how can you justify the inequality, since $\frac{\log x}{\sqrt{x}}$ only reaches its maximum until around $x\approx 7.39$, which is well above the $x>2$ condition stated in the question.
Eric Naslund's solution to post C states that 
$$\sum_{d> x}\frac{\mu(d)\log d}{d^2}=O(1/x)$$
but this less elementary result is presumably even harder to derive.
Can anyone identify the "correct" answer, and explain why?
 A: Hint:
$$
\sum_{d \geq x} \frac{\log d}{d^2} \leq \frac{\log x}{x^2} + \int_x^\infty \frac{\log t}{t^2}\,dt.
$$
A: One way to show that 
$$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}=O\left(\frac{\log x}{x}\right)$$
is to split into diadic intervals and use the bound $\sum_{n> y}\frac{1}{n^2}\ll \frac{1}{y}$.  We have that 
$$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}\leq\sum_{k=1}^{\infty}\sum_{2^{k-1}x<n\leq2^{k}x}\frac{\log n}{n^{2}} $$ 
$$\leq \sum_{k=1}^{\infty}\log(2^{k}x)\sum_{2^{k-1}x<n\leq2^{k}x}\frac{1}{n^{2}} $$ 
$$\ll \sum_{k=1}^{\infty}\log(2^{k}x)\frac{1}{2^{k-1}x}$$
$$=\frac{\log x}{x}\sum_{k=1}^{\infty}\frac{1}{2^{k-1}}+\frac{1}{x}\sum_{k=1}^{\infty}\frac{k\log2}{2^{k-1}}\ll\frac{\log x}{x}.$$
In my answer that you mentioned, I was thinking of this elementary bound, and forgot the $\log x$ term. However, the result I quoted is not false - we just need to take into account the cancellation resulting from the $\mu(n)$ term. Letting $M(x)=\sum_{n\leq x} \mu(n)$, integration by parts tells us that
$$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}=\int_{x}^{\infty}\frac{\log t}{t^{2}}d\left(M(t)\right)$$
$$=\frac{M(x)\log x}{x^{2}}+\int_{x}^{\infty}\frac{M(x)(2\log x-1)}{x^{3}}dx.$$ By the prime number theorem, there exists $c>0$ such that $$M(x)\ll xe^{-c\sqrt{\log x}},$$ and so we see that there is a constant $c_1>0$ such that $$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}\ll \frac{1}{x}e^{-c\sqrt{\log x}}.$$
