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 $\zeta(0) = -1/2$.
 A: 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. 
