Prove that $\sum_{n=1}^\infty \mu(n) \frac{\log n}{n} =1$ 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.
 A: One approach is to use Perron's formula.  In this case, we have that for $x\notin \mathbb{N}$
$$\sum_{n\leq x}\frac{\mu(n)}{n}\log n=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{-\zeta^{'}(s+1)}{\zeta(s+1)^{2}}\frac{x^{s}}{s}ds.$$ 
The residue of $\frac{-\zeta^{'}(s+1)}{\zeta(s+1)^{2}}\frac{x^{s}}{s}$ at $s=0$ is $1$, which leads to our expected answer.  (I am skipping the majority of the work which is bounding the integrand over the other contours, and moving things into the zero free region.)
A: 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 $$-\sum_{n\leq x}\frac{\mu(n)\log n}{n}=1+o(1).$$
A: If we modify the functional equation of $\zeta(s)$ we get
$$\cfrac{1}{\zeta(s)}= 2\,\cfrac{(2\pi)^{-s}\cos\frac{\pi s}{2}\;\Gamma(s)}{\zeta(1-s)}$$
Differentiating both sides yields by the quotient rule
$$\cfrac{\mathrm{d}}{\mathrm{d}s}\cfrac{1}{\zeta(s)} = 2\,\cfrac{\zeta(1-s)\cfrac{\mathrm{d}}{\mathrm{d}s}\big[(2\pi)^{-s}\cos\frac{\pi s}{2}\;\Gamma(s)\big]-(2\pi)^{-s}\cos\frac{\pi s}{2}\;\Gamma(s)\cfrac{\mathrm{d}}{\mathrm{d}s}\zeta(1-s)}{\zeta^2(1-s)}.$$
Now we need to know two things: $(uvw)' = u'vw+uv'w+uvw'$ and that $\zeta(0) = -\frac{1}{2}$, which can be easily proven by the functional equation and the fact that $(s-1)\zeta(s) = 1, \quad s\rightarrow1$. If we look even more closely we see that $\cos(\frac{\pi}{2}) = 0$ so we only have to evaluate one derivative.
$$\lim_{s\rightarrow1}\cfrac{\mathrm{d}}{\mathrm{d}s}\cfrac{1}{\zeta(s)} = 2\cfrac{-\frac{1}{2}\cdot (2\pi)^{-1}(-\frac{\pi}{2}\sin\frac{\pi\cdot1}{2})\,\Gamma(1)-0}{(-\frac{1}{2})^2} = 1$$
Have fun, Daniel.
