One can show that the Prime Number Theorem is equivalent to the statement
$$ A(x):= \sum_{n \leq x} \frac{\mu(n)}{n}=o(1),\qquad \tag1$$
i.e. that $A(x) \to 0$ as $x \to \infty$. Given that the identity
$$\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} \qquad\qquad \tag2$$
holds (elementary) for $\Re s >1$, line (1) seems to encode nothing more than the fact that $\zeta(s)$ is non-vanishing on the critical line $\Re s=1$ (equivalent to the PNT). By analogy, I would expect that
$$\sum_{n=1}^\infty \frac{\mu(n)\log n}{n^s} = \frac{d}{ds} \frac{1}{\zeta(s)}=\frac{-\zeta'(s)}{\zeta(s)^2} \qquad \tag3$$
to hold not only for $\Re s >1$, but for $\Re s =1$ (once again, using nothing more than the PNT). If so, then
$$\sum_{n=1}^\infty \frac{\mu(n) \log n}{n}=1 \tag4$$
using $(3)$ while ignoring the removable singularity at $s=1$. How can I back up this intuition?
I know that $(4)$ holds under the Riemann Hypothesis, but I believe (and would consequently like to show) that the PNT alone should suffice.
Best Answer
There is a nice, short elementary solution
By the identity $$\left(\frac{1}{\zeta(s)}\right)^{'}=-\frac{\zeta^{'}(s)}{\zeta(s)}\frac{1}{\zeta(s)},$$ we have that
$$-\sum_{n\leq x}\frac{\mu(n)\log n}{n}=\sum_{n\leq x}\frac{1}{n}\sum_{d|n}\mu(d)\Lambda\left(\frac{n}{d}\right).$$ This sum equals $$\sum_{dk\leq x}\frac{\mu(d)\Lambda\left(k\right)}{dk} =\sum_{dk\leq x}\frac{\mu(d)\left(\Lambda\left(k\right)-1\right)}{dk} +\sum_{dk\leq x}\frac{\mu(d)}{dk}.$$ The right most sum is easily seen to equal $1$, as $$\sum_{dk\leq x}\frac{\mu(d)}{dk}=\sum_{n\leq x}\frac{1}{n}\sum_{d|n} \mu(d)=1,$$ so we need only show that the other term is $o(1)$. By the hyperbola method $$\sum_{dk\leq x}\frac{\mu(d)\left(\Lambda\left(k\right)-1\right)}{dk}=\sum_{d\leq\sqrt{x}}\frac{\mu(d)}{d}\sum_{k\leq\frac{x}{d}}\frac{\left(\Lambda\left(k\right)-1\right)}{k}+\sum_{k\leq\sqrt{x}}\frac{\left(\Lambda\left(k\right)-1\right)}{k}\sum_{\sqrt{x}<d\leq\frac{x}{k}}\frac{\mu(d)}{d}.$$ Since $$\sum_{k\leq y}\frac{\left(\Lambda\left(k\right)-1\right)}{k}=o\left(\frac{1}{\log y}\right)\text{ and }\sum_{d\leq y}\frac{\mu(d)}{d}=o\left(\frac{1}{\log y}\right)$$ by the prime number theorem, the above is $$\ll o\left(\frac{1}{\log x}\right)\left(\sum_{d\leq\sqrt{x}}\frac{1}{d}\right)\ll o\left(1\right),$$ and so we see that $$-\sum_{n\leq x}\frac{\mu(n)\log n}{n}=1+o(1).$$