When I tested this in Mathematica, I had expected it to say it did not converge. However, I got this:

$$\prod_{n=1}^\infty n^{\mu(n)}=\frac{1}{4 \pi ^2}$$

Note: this is the reciprocal of (3) zeta-regularized product over all primes.

This indicates there is a slight skew toward the square-free numbers that have an odd number of factors.

I haven't been able to find this in the literature, but my guess is that it is already known.

Can someone point me in the right direction?

Edit: We have been told it is a bug. Mathematica V.10 does not produce a symbolic answer.

Edit 2 We can make this work by creating a function $f(k)$ that returns the $k$-th square-free number (in the constructive? order). We can find a few ideas from A019565.


$$\sum_{n=1}^\infty \mu(f(n))$$

At the steps when $n>2$ is a power of $2$, sum$=0$ and prod$=1$.
Also, at those steps, $f(n)$ is the primorial with $p_k$ as the greatest factor, with $k=\frac{\log(n)}{\log(2)}$

There is a revised question over here. It is a variation to Edit 2.

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    $\begingroup$ There's no way this product can converge, as its terms don't tend to 1. Goodness knows what Mathematica is trying to say with its "evaluation" of the product. $\endgroup$ Commented Mar 12, 2013 at 12:08
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    $\begingroup$ I have posted at mathematica.stackexchange.com/q/21149/973 to get some info on whether this is correct or not. $\endgroup$ Commented Mar 12, 2013 at 12:34
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    $\begingroup$ Doesn't Mathematica also silently apply summation methods to divergent series as well? Why declare this a bug, but not all of those? $\endgroup$ Commented Mar 12, 2013 at 17:07
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    $\begingroup$ @Gerald Edgar: the edit seems like a consequence of the follwing comment on mathematica.SE "I'm told it's a bug and will be treated as such. – Daniel Lichtblau". $\endgroup$
    – user9072
    Commented Mar 12, 2013 at 18:50
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    $\begingroup$ In its current state, I don't see where there's a question. $\endgroup$ Commented Mar 26, 2013 at 11:29

3 Answers 3


So of course it does not converge. The behavior is interesting however. Below is the graph of the $\ln$ of the product going up to $5000.$ enter image description hereThere can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. At $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simple convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. The partial product $\prod_{n=1}^N n^{\mu(n)}$ never has an even denominator for $5 \lt N \le 10000$ . However the numerator is just four times an odd number for $N=6590$ and $N=6593.$ So, if forced to guess, I'd guess that it is odd someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.

  • $\begingroup$ Image not loading. $\endgroup$
    – VS.
    Commented Apr 17, 2020 at 19:38
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    $\begingroup$ OK. I restored it. $\endgroup$ Commented Apr 19, 2020 at 5:56

We have $$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$ Taking the derivative with respect to $s$, we get the following $$- \frac{\zeta'(s)}{\zeta(s)^2} = - \sum_{n=1}^{\infty} (\log n) \frac{\mu(n)}{n^s} \quad \mbox{for} \ Re(s)>1.$$

If we plug in $s=0$ to the second equation (which is not allowed, because the second equation is only valid for $Re(s)>1$) we get $$ - \frac{\zeta'(0)}{\zeta(0)^2} = - \sum_{n=1}^{\infty} \mu(n) \log(n) \quad \mbox{(FALSE EQUATION)}.$$

We have $$\zeta(0) = - \frac{1}{2} \quad \zeta'(0) = - \frac{1}{2} \log(2 \pi)$$ so $$ - \frac{\zeta'(0)}{\zeta(0)^2} = 2 \log(2 \pi)$$

So, if we believed the false equation, we would have $$\sum_{n=1}^{\infty} \mu(n) \log(n) = -2 \log (2 \pi) \ \mbox{and} \ \prod_{n=1}^{\infty} n^{\mu(n)} = \frac{1}{4 \pi^2}.$$

I don't know why Mathematica thinks it's okay to plug in $s=0$.

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    $\begingroup$ @Speyer: "I don't know why Mathematica thinks it's okay to plug in $s=0$." Secretly Mathematica is a physicist? $\endgroup$ Commented Mar 12, 2013 at 15:18
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    $\begingroup$ Well we expect the formula for $-\zeta'/\zeta^2$ to hold for ${\rm Re}(s) > 1/2$... But certainly not for $s=0$. $\endgroup$ Commented Mar 12, 2013 at 15:20
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    $\begingroup$ One way to regularize $\sum_{n=1}^N f(n)$ is (from terrytao.wordpress.com/2010/04/10) by taking a cutoff function $\eta$ with compact support satisfying $\eta(0) = 1$, and using $\sum_{n=1}^\infty \eta(n/N) f(n)$. For $\eta = \chi_{[0,1]}$, you get the ordinary sum, and for $\eta(x) = (1-x)\chi_{[0,1]}$ you get Cesáro. If you want a good asymptotic expansion, $\eta$ should be smooth, or else boundary effects crop up. At any rate, you get your value of $2 \log(2\pi)$ plus $\sum_\rho c_\rho N^\rho$ as $\rho$ varies over zeroes of $\zeta'$, which I don't know how to control. $\endgroup$
    – S. Carnahan
    Commented Mar 13, 2013 at 2:14
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    $\begingroup$ @David Speyer, Cesaro summability cannot be true because in that case we would have an analytic continuation of $\zeta'/\zeta^2$ on $\textrm{Re}(s)>0$. $\endgroup$ Commented Mar 18, 2013 at 3:12
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    $\begingroup$ @Jason Starr Mathematica uses formal series, that is why this answer is ok for it. $\endgroup$ Commented Dec 14, 2013 at 2:47

This is a discussion following from David Speyer about Cesaro summability of the series. This series is not Cesaro summable in any order.

In "The General Theory of Dirichlet Series" by Hardy and Riesz. Chapter 4.

Let $\kappa\geq 0$. $0\leq \lambda _0<\lambda_1<\cdots<\lambda_n\rightarrow\infty$, and write $$ A_{\lambda}^{\kappa}(w) = \sum_{\lambda_n \textrm{<}w}(w-\lambda_n)^{\kappa}a_n. $$

\bf Summability $(\lambda,\kappa)$. \rm If $A_{\lambda}^{\kappa}(w)\sim Aw^{\kappa}$ as $w\rightarrow\infty$, then we say that $\sum a_n$ is summable $(\lambda,\kappa)$ to sum $A$.

The second consistency theorem(we do not need first consistency theorem here) states that If $\sum a_n$ is summable $(l,\kappa)$ where $l_n=e^{\lambda_n}$, then it is summable $(\lambda,\kappa)$ to the same sum. In particular, Cesaro summability of order $\kappa$ implies $(\log n, \kappa)$ summability to the same sum.

We consider the formula $$ \sum_{n=1}^{\infty}a_ne^{-\lambda_ns}=\frac{1}{\Gamma(\kappa+1)}\int_0^{\infty}s^{\kappa+1}e^{-s\tau}A_{\lambda}^{\kappa}(\tau)d\tau.$$

This is originally valid if $\sigma> max(0,\sigma_c)$ where $\sigma_c$ is the abscissa of convergence of the left series. On the other hand, the right allows an analytic continuation of the function represented by the series on the left up to $\sigma>0$.

What we use here is $a_n = \mu (n)\log n$ and $\lambda_n = \log n$ .

Hence, assuming Cesaro summability of any order $\kappa$, will allow $$-\frac{\zeta'(s)}{\zeta^2(s)}$$ to have an analytic continuation to $\sigma>0$ which contradicts the fact that $\zeta(s)$ has zeros on the critical line.


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