# Regularized sums of Mobius sequence

Do $\lim_{s \rightarrow \infty} \sum_{n \geq 1} \mu(n) e^{-n/s}$ and $\lim_{s \rightarrow \infty} \sum_{n \geq 1} \mu(n) e^{-n^2/s^2}$ both equal $-2$?

Experimentally this seems plausible (up through $s=10^6$).

On a related theme, does the Dirichlet series $\sum_{n \geq 1} \mu(n) n^{-s}$ converge to $1/\zeta(s)$ for all real $s$ between 0 and 1? If so, then sending $s \rightarrow 0^+$ would assign the divergent sum $\sum_{n \geq 1} \mu(n)$ the same regularized value as the first two regularization procedures, since $\zeta(0) = -1/2$.

• Both of those are $\Omega_{\pm}(s^{1/2})$ as $s\rightarrow\infty$. – Sungjin Kim Apr 5 '18 at 23:23

No! This is very badly false -- the Riemann zeta function has non-trivial zeros. For example, suppose that $M(x) = \sum_{n=1}^{\infty} \mu(n) e^{-nx}$ tends to $-2$ as $x\to 0$ (I've rewritten your hypothesis with $x=1/s$). In particular, you're assuming that $M(x)$ is always bounded. But in that case note that $$\int_0^{\infty} M(x) x^{s}\frac{dx}{x} = \sum_{n=1}^{\infty}\mu(n) \int_0^{\infty} e^{-nx} x^{s} \frac{dx}{x} = \frac{\Gamma(s)}{\zeta(s)},$$ where the integral a priori converges for Re$(s)>1$ (because as $x\to \infty$ clearly $M(x)\ll e^{-x}$ decreases exponentially, and trivially as $x\to 0^+$ we can use $|M(x)| \ll x^{-1}$). The assumption that $M(x)$ is bounded as $x\to 0^+$ now implies that the integral actually makes sense in Re$(s)>0$, or in other words that $\zeta(s)$ has no zeros! See also my answer to Is it possible to show that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges? .
Alternatively, one can write down an explicit formula for $M(x)$ in terms of zeros of $\zeta(s)$. Namely for some $c>1$ $$M(x) = \frac{1}{2\pi i} \int_{c- i\infty}^{c+i\infty} \frac{1}{\zeta(s)} x^{-s} \Gamma(s) ds,$$ and moving the line of integration to the left, we find $$M(x) = \sum_{\rho} \frac{\Gamma(\rho)}{\zeta^{\prime}(\rho)} x^{-\rho}+ \frac{1}{\zeta(0)} + O(x),$$ where the sum is over non-trivial zeros of zeta (assumed to be simple for convenience). The second term above arises from the pole of the Gamma function at $s=0$, and note that it equals $-2$. The error term can be made explicit in terms of the poles at $-1$, $-2$, etc. This shows why $M(x)$ will have to be of size at least $x^{-1/2}$ occasionally, and further explains why the numerical evidence is misleading: the first zero of $\zeta$ has large ordinate (about $14.1\ldots$), and $\Gamma(1/2+14.1\ldots i)$ is very small in size.
• I'm sure I'm missing something, but why does $\int_0^\infty x^s\,\frac{dx}{x}$ converge "a priori" for Re$(s)>1$? For, say, $s=2$, certainly $\int_0^\infty x\,dx$ doesn't converge. – Joe Silverman Mar 1 '15 at 23:33
• For anyone who's interested in knowing just how small $\Gamma(1/2+14.1\dots i)$ is, Mathematica reports that its real and imaginary parts are on the order of $10^{-10}$. – James Propp Mar 2 '15 at 0:28
• @JamesPropp: By Stirling's formula, $\Gamma(1/2+it)$ is much like $\sqrt{2\pi}e^{-\pi t/2}$, so one can check the $10^{-10}$ order even with a simple calculator. – GH from MO Mar 2 '15 at 10:05
• @JamesPropp: From Lucia's response it follows that $M(x)$ is the inverse Mellin transform of $\Gamma(s)/\zeta(s)$. Hence, by the residue theorem, $M(x)$ can be approximated well by a large finite sum of the sum of residues of $x^{-s}\Gamma(s)/\zeta(s)$ at the zeta-zeros. So I don't think there is an asymptotic expansion of $M(x)$, it is more like an infinite sum of winding terms each of size $x^{-1/2}$ and its fluctuations are nontrivial to analyze. This is a similar quantity as $\pi(x)-\mathrm{li}(x)$ or $\psi(x)-x$ in prime number theory. It would be interesting to explain your $-2$ though. – GH from MO Mar 2 '15 at 16:23