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.


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

  • $\begingroup$ I apologize for commenting so late after this post, but would you mind explaining how the estimates $\sum_{k\leq y}\frac{\Lambda(k)-1}{k}=o(1/\log y)$ and $\sum_{d\leq y}\frac{\mu(d)}{d}=o(1/\log y)$ follow from the prime number theorem? $\endgroup$ Nov 11 '16 at 18:25
  • $\begingroup$ @ColinDefant For that - with a small correction, $\sum_{k \leqslant y} \frac{\Lambda(k)-1}{k} = -2\gamma + o(1/\log y)$, but the constant is harmless - one needs the PNT with sufficiently good (but much weaker than the proven) error bounds [at least, I don't know how to prove it without, and neither did Landau]. The plain PNT without error bounds, i.e. just $\sum_{n = 1}^{\infty} \frac{\mu(n)}{n}$ converges, or $\psi(x) \sim x$, or $\pi(x) \sim x/\log x$, or …, is equivalent to $$\sum_{n \leqslant x} \frac{\mu(n)\log n}{n} \in o(\log x).$$ $\endgroup$ Mar 10 '18 at 19:34
  • $\begingroup$ @DanielFischer, the question seems to be implying the sum goes to +1. The answer seems to be deriving -1. I suppose there a typo in the question. (Writing to you as you seem to have moderator access and may be able to correct it. Thanks!) $\endgroup$
    – Shree
    Mar 16 at 17:12
  • $\begingroup$ Link to Landau's book (relevant page is 569 books.google.com/… $\endgroup$
    – Shree
    Mar 16 at 18:23

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.)

  • $\begingroup$ 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 $c$? $\endgroup$
    – awwalker
    Dec 12 '12 at 18:21
  • $\begingroup$ @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. $\endgroup$ Dec 12 '12 at 18:32

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.

  • $\begingroup$ This is hardcore! $\endgroup$
    – TravorLZH
    Nov 22 '20 at 15:57
  • $\begingroup$ @TravorLZH It is "overkill" but not a solution. That $(1/\zeta(s))'$ is $1$ at $s = 1$ can be shown in a far simpler way: $\zeta(s)$ has a simple pole at $s = 1$ with residue $1$, so $\zeta(s) = 1/(s-1) + $ higher order terms near 1. Thus $1/\zeta(s) = (s-1) + $ higher powers of $s-1$, so $(1/\zeta(s))' = 1 + O(s-1)$ near 1. Hence $(1/\zeta(s))'|_{s=1} = 1$. And $(1/\zeta(s))' = -\sum_{n \geq 1}\mu(n)\log(n)/n^s$ for ${\rm Re}(s) > 1$, so if $\sum_{n \geq 1} \mu(n)\log(n)/n$ converges then the value is $-1$ (Abel's theorem for Dirichlet series), but that does not prove the sum is $1$. $\endgroup$
    – KCd
    Apr 17 at 18:23
  • $\begingroup$ A more basic example is $1/\zeta(s)$ at $s = 1$. The simple pole of $\zeta(s)$ at $s = 1$ implies a simple zero of $1/\zeta(s)$ at $s = 1$. Since $1/\zeta(s) = \sum_{n \geq 1} \mu(n)/n^s$ for ${\rm Re}(s) > 1$, if $\sum_{n \geq 1} \mu(n)/n$ converges then the value is $0$ (Abel's theorem for Dirichlet series), but that does not prove $\sum_{n \geq 1} \mu(n)/n = 0$. In fact, "$\sum_{n \geq 1} \mu(n)/n = 0$" is known to be equivalent to the Prime Number Theorem ($\pi(x) \sim x/\log x$, or equivalently $\psi(x) \sim x$): each implies the other by methods far simpler than proving PNT. $\endgroup$
    – KCd
    Apr 17 at 18:30

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