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Denote by $\mu$ the Mobius function. It is known that for every integer $k>1$, the number $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.

Letting $k\rightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form

$$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=0,$$

since the probability that an integer is ``$1$-free'' is zero ?

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    $\begingroup$ It is true that the PNT is equivalent to $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$. It is also relatively easy to prove that $\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = 0$. The hard part is proving that $\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \lim_{x \to \infty} \sum_{n \leq x} \frac{\mu(n)}{n}$. This is highly nontrivial! $\endgroup$
    – Peter Humphries
    Commented Apr 20, 2019 at 21:48
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    $\begingroup$ In general, limit of sums of series $\neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT. $\endgroup$
    – Wojowu
    Commented Apr 20, 2019 at 21:48
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    $\begingroup$ I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there. $\endgroup$
    – Gro-Tsen
    Commented Apr 20, 2019 at 22:41
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    $\begingroup$ I agree with Fourton and have voted accordingly $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2019 at 22:43
  • $\begingroup$ @YemonChoi, is that a new nickname for @‍Gro-Tsen? :-) $\endgroup$
    – LSpice
    Commented Apr 22, 2019 at 1:03

1 Answer 1

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You ask:

Denote by $\mu$ the Mobius function. It is known that for every integer $k>1$, the number $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.

Letting $k\rightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form

$$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=0,$$

since the probability that an integer is ``$1$-free'' is zero ?

As pointed out by the users @wojowu and @PeterHumphries, it is true that the PNT is equivalent to

$$\lim_{x \to \infty} \sum_{n\leq x} \frac{\mu(n)}{n}=0,$$ and it is relatively easy to prove that

$$\lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}=0.$$ The real difficulty lies in proving that

$$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{\mu(n)}{n}= \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s},$$ which is highly nontrivial and requires intricate arguments.

In particular, as pointed out by @TerryTao in the comments:

if $t\neq 0$ is real, then

$$ \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{n^{it}}{n^s},$$

can be shown to converge to a finite value, whereas $$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{n^{it}}{n}$$

is undefined. So at a bare minimum one has to somehow stop $\mu(n)$ from "pretending" to be like $n^{it}$. This turns out to be basically equivalent to preventing $\zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.

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    $\begingroup$ Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates... $\endgroup$
    – Nate Eldredge
    Commented Apr 21, 2019 at 3:03
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    $\begingroup$ @NateEldredge Or magic guarantees like (weak) compactness... $\endgroup$
    – Yemon Choi
    Commented Apr 21, 2019 at 3:16
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    $\begingroup$ In particular, if $t$ a non-zero real, then $\lim_{s \to 1^+} \sum_{n=1}^\infty \frac{n^{it}}{n^s}$ can be shown to converge to a finite value, whereas $\lim_{x \to \infty} \sum_{n \leq x} \frac{n^{it}}{n}$ is undefined. So at a bare minimum one has to somehow stop $\mu(n)$ from "pretending" to be like $n^{it}$. This turns out to be basically equivalent to preventing $\zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem. $\endgroup$
    – Terry Tao
    Commented Apr 21, 2019 at 17:49
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    $\begingroup$ @TerryTao Vague question: Why does it suffice, to prove the prime number theorem, to show that $\mu(n)$ does not pretend to be like $n^{it}$? In other words, why is that pretention the only obstruction? $\endgroup$
    – mathworker21
    Commented Apr 24, 2019 at 3:15
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    $\begingroup$ The reason is spectral theory - traditionally expressed via complex analysis (e.g. Perron's formula), but also expressible via Fourier analysis (e.g. Parseval formula) or from the Gelfand theory of Banach algebras. One can for instance start from the reproducing formula $\mu(n) \log n = - \sum_{d|n} \mu(d) \Lambda(n/d)$ (traditionally Selberg's symmetry formula is used instead). On the Fourier side, this type of formula can be used to show $\Lambda$ either has a "large Fourier coefficient" ($\mu$ pretends to be $n^{it}$) or $\mu$ has mean zero; this is a special case of Halasz's theorem. $\endgroup$
    – Terry Tao
    Commented Apr 24, 2019 at 4:26

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