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Oct 31, 2011 at 10:55 vote accept Kevin Smith
Oct 31, 2011 at 7:37 comment added Kevin Smith $M(n)$ is Mertens function (the last term counted with weight $1/2$).
Oct 31, 2011 at 0:20 comment added Frank Thorne Sorry, what is $M(n)$?
Oct 30, 2011 at 23:41 answer added Greg Martin timeline score: 3
Oct 30, 2011 at 23:34 comment added Kevin Smith One can see that it is not of constant because $M(x)$ is known to be zero infinitely often.
Oct 30, 2011 at 23:23 comment added Kevin Smith Error: should read $c_n=2\mu(n)M(n)$.
Oct 30, 2011 at 23:21 comment added Kevin Smith Thanks for the input, Frank. I am interested in the sequence $c_n=2\mu{n}M(n)$- its partial sum is $M^2(x)$ - and so satisfies the statement I made in the question. I don't expect it to be of constant sign for sufficiently large $n$ (that would be asking a lot!), yet I am wondering if the pole on the real axis (which occurs at $2\Omega$, where $\Omega$ is the largest real part of a zero of $\zeta(s)$) + some other hypotheses would imply the positivity of the partial sum.
Oct 30, 2011 at 20:24 comment added Frank Thorne As Daniel explained, you can rig counterexamples to any simple statement you might hope to prove. However, theorems of this flavor tend to be provable under conditions on the $c_n$ which are strong, but typical for the kinds of Dirichlet series that show up in number theory. What example are you interested in?
Oct 30, 2011 at 19:33 answer added Daniel Loughran timeline score: 4
Oct 30, 2011 at 18:57 history edited Kevin Smith CC BY-SA 3.0
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Oct 30, 2011 at 18:51 history asked Kevin Smith CC BY-SA 3.0