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One can show that the Prime Number Theorem is equivalent to the statement $$ A(x):= \sum_{n \leq x} \frac{\mu(n)}{n}=o(1),\qquad \qquad (1)$$ i.e. that $A(x) \to 0$ as $x \to \infty$. Given that the identity $$\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} \qquad\qquad \qquad (2)$$ 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 $$\sum_{n=1}^\infty \frac{\mu(n)\log n}{n^s} = \frac{d}{ds} \frac{1}{\zeta(s)}=\frac{-\zeta'(s)}{\zeta(s)^2} \qquad \qquad (3)$$ to hold not only for $\Re s >1$, but for $\Re s =1$ (once again, using nothing more than the PNT). If so, then $$\sum_{n=1}^\infty \frac{\mu(n) \log n}{n}=1 \qquad \qquad \quad(4)$$ 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.

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2 Answers 2

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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 $$-\sum_{n\leq x}\frac{\mu(n)\log n}{n}=1+o(1).$$

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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.)

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This requires $\mu(n) \log n/n^s$ to be absolutely convergent for all $\Re s >c-1$, with $c < 0$ (to isolate the residue at $\zeta(1)$). This is a result that is currently conditional. To clarify, what is your $c$? –  A Walker Dec 12 '12 at 18:21
    
@A Walker: The value of $c$ is greater than $0$. However we can look at a contour integral in the plane which contains the line segment $c-iT$, $c+iT$, and move the contour past $c=0$, into the zero free region without picking up any additional residues. $\zeta(s+1)$ is a meromorphic function with a pole at only $s=0$, so we are free to move the contour where we like, noting that we must be careful since the zeros will have an effect as well. This idea is essentially the idea behind the proof of the quantitative prime number theorem. –  Eric Naslund Dec 12 '12 at 18:32

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