I read the statement that

$$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \textrm{ for } \Re(s) > 1 \qquad (*)$$

In fact I can guess what the proof is: just expand both $\zeta$ and the right hand side of (*) as an Euler product, use $\Re(s) > 1$ to handwave away any concerns of changing the order of taking limits and taking products and conclude that the outcome of multiplying the right hand side of (*) by $\zeta(s)$ is the product of infinitely many $1$'s, hence 1 itself. So far so good.

Now suppose I wanted to evaluate the the limit $s \to 1^+$ of the right hand side of ( * ) , or in layman's terms, compute the sum $\sum_{n=1}^\infty \frac{\mu(n)}{n}$. From (*) I would quickly conclude that

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

since $\zeta(1) = \sum_{n=1}^\infty \frac{1}{n} = \infty$ (by the standard 'powers of 1/2'-argument) and, as we all know, $1/\infty = 0$. Very nice, very elementary, or... so it seems.

However the statement (**) is far from elementary: it is equivalent to the prime number theorem! (If you find that equivalence surprising: so did I. I asked a question about it three years ago, but the current question is about a different type of surprise that arises if we take this equivalence as given.)

We all know that the PNT is hard to prove. So what is wrong with simply taking the the limit $s \to 1$ in (*) as I did above? Am I secretly interchanging two limits where that is not allowed?

Where am I oversimplifying things?

I have a very vague feeling what is going on here (but perhaps I am completely off, so correct me if I am wrong) and that is that there is some weird theorem lurking in the background that states that we can only extend the equation (*) to the point $s = 1$ if and only if we can extend it to the entire line $\{s \colon \Re(s) = 1\}$. But what kind of weird theorem would that be? It goes again my nearly lifelong experience that one can do mathematics just fine without even realizing that complex numbers off the real line exist.

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    $\begingroup$ You are claiming that $\lim_{s \rightarrow 1^+}\,\sum_{n=1}^{\infty}{\frac{\mu(n)}{n^s}}=\sum_{n=1}^{\infty}{\lim_{s \rightarrow 1^+}\,\frac{\mu(n)}{n^s}}$. This is interchanging limits, is it not? $\endgroup$
    – Aphelli
    Apr 18 at 21:13
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    $\begingroup$ Hm, yes, but why is it forbidden? O wait, I can see it, the sum on the right hand side converges, but it does not converge absolutely because replacing the $\mu$'s with $|\mu|$'s would make the sum run off to infinity. Hmmmm.... Maybe it is that simple? $\endgroup$
    – Vincent
    Apr 18 at 21:17
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    $\begingroup$ That’s precisely it. It takes careful justification to interchange limits, any limits (including convergent – or even absolutely convergent – sums). $\endgroup$
    – Aphelli
    Apr 18 at 21:23
  • $\begingroup$ $$\lim_{s\to1^+}\sum_{n}\frac{\mu(n)}{n^s}=\lim_{s\to1^+}\prod_p\left(1-\frac1{p^s}\right)$$ Thaking the log of right side: $$\lim_{s\to1^+}\sum_p\log\left(1-\frac1{p^s}\right)$$ and $\log(1-x)<-x$ for $x>0.$ So $$\leq-\lim_{s\to 1^+}\sum_{p}\frac1{p^s}$$ $\endgroup$ Apr 18 at 21:27
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    $\begingroup$ the fact that $\lim_{s \rightarrow 1^+}\,\sum_{n=1}^{\infty}{\frac{\mu(n)}{n^s}}=0$ is trivial given the usual manipulations that show that $\sum_{n=1}^{\infty}{\frac{\mu(n)}{n^s}}=1/\zeta(s), \Re s >1$; the hard part is to show that one can interchange limits and series as that is not true in general (for example for any $t \ne 0$, one has $\frac{\zeta'}{\zeta}(\sigma+it) \to \frac{\zeta'}{\zeta}(1+it), \sigma \to 1^+$ but the corresponding Dirichlet series oscillates with bounded partial sums $\endgroup$
    – Conrad
    Apr 18 at 22:00

1 Answer 1


The question in the title is:

$\lim_{s \to 1^+} 1/\zeta(s) = 0$ obvious or not?

From the body of the text, I'm not sure that this is the real question the poster intended, but I answer the above question as it stands.

Claim: Suppose $a_n$ is a decreasing sequence of positive real numbers with $a_n\to 0$ and $\sum_{n=1}^{\infty} a_n = +\infty.$ Then, $\lim_{s\to 1^+} \sum_{n=1}^{\infty} {a_n}^s = \infty.$

Proof: Let $\gamma\in\mathbb{R}_{>0}$ (think about $\gamma$ being a large positive real number). We are given that $\sum a_n$ diverges to $+\infty.$ So, $\exists k$ such that $\sum_{n=1}^{k} a_n > 2\gamma$.

Proposition: $\exists s>1$ such that ${a_n}^s > a_n/2\quad \forall n\in \{1,\ldots,k\}.$ Proof of proposition: Take $s = 1+a_k.$

Therefore, $\ \sum_{n=1}^{\infty}{a_n}^s > \sum_{n=1}^{k}{a_n}^s > \sum_{n=1}^{k}{\frac{a_n}{2}} > \gamma.$

Furthermore, $1 <s' < s \implies \sum {a_n}^{s'} > \sum{a_n}^s,\ $ and so $\lim_{s\to 1^+} \sum_{n=1}^{\infty} {a_n}^s = \infty.$

This proves that claim. Now take $a_n = \frac{1}{n}.$ We see that the question in the title is proven.

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    $\begingroup$ The series in OP's question is not a positive series. $\endgroup$ Apr 18 at 23:21
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    $\begingroup$ @SangchulLee I thought $\zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^3} + \ldots$ $\endgroup$ Apr 18 at 23:23
  • $\begingroup$ I assume Adam is applying the claim of this answer to $\zeta(s)$, not $1/\zeta(s)$. $\endgroup$
    – runway44
    Apr 18 at 23:25
  • $\begingroup$ Yeah but in what universe is $\lim_{s\to 1^+} \frac{1}{f(s)} = 0$ not that same as: $\lim_{s\to 1^+} f(s) = +\infty$ ? $\endgroup$ Apr 18 at 23:28
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    $\begingroup$ Oh, I see. I thought it was a question about obtaining $(**)$ from $(*)$, as it was the main topic in the comments. Indeed, showing $\zeta(s)\to\infty$ as $s\to1^+$ should be no hard, for instance using the monotone convergence theorem. $\endgroup$ Apr 18 at 23:29

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