TITLE: mertens-function in the light of divergent summation - what summation method were best adapted
QUESTION [5 upvotes]: Just reading about the Mertens-function in the other thread 
Mertens function I remember an earlier attempt to apply divergent summation
to the series which is constructed of the Moebius-function 
at consecutive arguments, or in other words of which the Mertens-function-values represent 
the partial sums.
Eulersummation, although relatively poorly adapted, suggested that the (divergent) sum should be 
meaningfully evaluated to -2. But that sequence of partial sums (although seldom exceeding only the squareroot of its current index) seems to be a specific difficult case for
such common summation methods - the approximation is relatively poor even for 128 terms. 
I tried Nörlund-means/Cesaro-sum, Euler-sums of different orders and also a selfmade matrix summation-method using the eulerian numbers (with which I could -on the other hand- well handle the even strongly 
diverging $ 0!-1!+2!-3!+...$ -series), but I tried not yet for instance Abel and Borel.
Q1: What method would be most appropriate to sum the series
$ S = \sum _{k=1}^{\inf} moebius(k)  $ 
Evaluation of 128 terms (Euler,Cesaro) suggested the result $S = -2$
Q2: And how could it be determined whether the Cesaro- and/or Euler-summation are at all capable to evaluate
that series to a final value?
Here is  of the summation.

REPLY [8 votes]: Well,
$$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac1{\zeta(s)}$$
for $s>1$, so setting $s=0$ should give
$$\sum_{n=1}^\infty\mu(n)=\frac1{\zeta(0)}=-2$$
as $\zeta(0)=-1/2$. :-)
I should add that this is a trick often used in analytic
number theory (for instance in Eisenstein series). More generally
given a divergent sum
$$S=\sum_{i\in I}a_i$$
then consider, for an appropriate choice of weights $b_i>0$
the series
$$f(s)=\sum_{i\in I}\frac{a_i}{b_i^s}.$$
We hope this converges in a suitable half-plane
and can be analytically continued to $0$. Then we "define"
$S=f(0)$.