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 a plot of the summation.

  • $\begingroup$ Can you try your methods on $\sum_{k=1}^\infty\mu(2k)$? or other arithmetic progressions? $\endgroup$
    – joro
    Commented May 28, 2012 at 9:30
  • $\begingroup$ 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? $\endgroup$ Commented May 28, 2012 at 11:03
  • $\begingroup$ 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? $\endgroup$
    – joro
    Commented May 28, 2012 at 11:30
  • $\begingroup$ That would surprise me - considering using $\mu(2k)$ plus $\mu(2k+1)$ should be (eulerian-) summable to a finite value... $\endgroup$ Commented May 28, 2012 at 12:16
  • 1
    $\begingroup$ Related: mathoverflow.net/q/438841/25104 $\endgroup$ Commented Jan 18, 2023 at 19:12

1 Answer 1


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" $S=f(0)$.

  • $\begingroup$ " 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? $\endgroup$ Commented Nov 27, 2010 at 9:51
  • $\begingroup$ @Gottfried: Generally, series with erratic sign patterns are not very amenable to series transformations. That series for reciprocal zeta is a commonly cited example. $\endgroup$ Commented Nov 27, 2010 at 14:26
  • $\begingroup$ @J.M. - yes, thanks too. I think I should reread the related chapters in my Knopp-copy. I found that passages, where the range of applicability was discussed, fairly difficult. I've got some more experience over the last few years and might be able to catch it up now easier... $\endgroup$ Commented Nov 27, 2010 at 17:33
  • $\begingroup$ Just came back to this thread by another discussion in sci.math. I'm setting the accept mark because Q2 may be to broad to be answered here, and I think the bot would reactivate the question otherwise. (I've taken Robin's solution/concept to my heart :-) ) $\endgroup$ Commented Jan 7, 2011 at 7:57

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