# mertens-function in the light of divergent summation - what summation method were best adapted

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.

• Can you try your methods on $\sum_{k=1}^\infty\mu(2k)$? or other arithmetic progressions?
– joro
May 28, 2012 at 9:30
• I can't make naive Euler/Cesaro - etc sums converge, using up to 200 terms. I'll see later, whether I can improve this. Can't we derive something in the spirit of Robin Chapman's answer? May 28, 2012 at 11:03
• Gottfried, I suspect it may be infinity using some methods -- naiively using Robin's answer (this may be wrong) I got 1/0 -- the Dirichlet series vanishes at zero... I suppose $\mu(2k+1)$ should converge in some sense though?
– joro
May 28, 2012 at 11:30
• That would surprise me - considering using $\mu(2k)$ plus $\mu(2k+1)$ should be (eulerian-) summable to a finite value... May 28, 2012 at 12:16
• Hm, you are probably right. mu(2k) has positive bias and mu(2k+1) negative to 10^6.
– joro
May 28, 2012 at 13:07

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" $S=f(0)$.
• " as $\zeta (0)= - 1/2$ " ... <arrggh>. I should have seen this, thank you! Well, the second question has more the character of exercising with the characteristics of that summation-procedures. In G.H.Hardy's and K.Knopp's books which cover the divergent series we find "not too strong diverging" series which still can not be Euler-summed - it seems mostly because of non-simpliness of their pattern of changing sign. That Moebius-series seems to be a specifically nasty one: even non-periodic. Is this the main reason? Nov 27, 2010 at 9:51