mertens-function in the light of divergent summation - what summation method were best adapted - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:22:06Z http://mathoverflow.net/feeds/question/47469 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47469/mertens-function-in-the-light-of-divergent-summation-what-summation-method-were mertens-function in the light of divergent summation - what summation method were best adapted Gottfried Helms 2010-11-26T23:15:54Z 2012-01-28T22:21:40Z <p>Just reading about the Mertens-function in the other thread <a href="http://mathoverflow.net/questions/47436/is-there-an-elementary-proof-that-the-mertens-function-is-not-ox-theta-if" rel="nofollow">Mertens function</a> 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.</p> <p>Eulersummation, although relatively poorly adapted, suggested that the (divergent) sum should be meaningfully evaluated to <em>-2</em>. 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 <em>128</em> 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.</p> <p>Q1: What method would be most appropriate to sum the series $S = \sum _{k=1}^{\inf} moebius(k)$ </p> <p>Evaluation of <em>128</em> terms (Euler,Cesaro) suggested the result $S = -2$</p> <p>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?</p> <p>Here is <a href="http://go.helms-net.de/math/images/mertenssum.png" rel="nofollow">a plot</a> of the summation.</p> http://mathoverflow.net/questions/47469/mertens-function-in-the-light-of-divergent-summation-what-summation-method-were/47489#47489 Answer by Robin Chapman for mertens-function in the light of divergent summation - what summation method were best adapted Robin Chapman 2010-11-27T08:13:02Z 2010-11-27T08:13:02Z <p>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 $\zeta(0)=-1/2$. :-)</p> <p>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" $S=f(0)$.</p>