Convergence of sum involving Mobius function (with Riemann Zeta)

  • Thread starter The_Shape
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Hello everyone.

I was hoping someone could clarify this "heuristic" argument I found online. First, what is the analytic function they speak of and is its derivative difficult to compute? Second, does this look like a legit argument? :

If you take the derivative w.r.t s of both sides of sum mu(n)/n^s=1/zeta(s) and have s->1, you get that the sum [mu(n)/n^s]* (-log n) is -zeta'(s)/(zeta(s))^2. Because you probably know that zeta(s)=1/(s-1)+analytic function, zeta'(s)=-1/(s-1)^2+analytic, so just by dividing, zeta'(s)/(zeta(s))^2 is [-1/(s-1)^2 + analytic]/[1/(s-1)^2+terms with 1/(s-1) and analytic], so multiplying by (s-1)^2/(s-1)^2 yields that as s->1, everything is 0 in numerator and denominator except for the leading terms, i.e. [-1+terms tending to zero]/[1+terms tending to zero], so it all cancels to -1. So the sum [mu(n)/n^s]* log(n) tends to -1 as s->1, so that again motivates that the sum mu(n)/n log(n) could well be -1.
 
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Answers and Replies

  • #2
mathman
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Piece of advice. Learn to write more clearly. Short sentences for each step would be helpful.
 
  • #3
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Uh, ok. Thanks for the advise. I copied this from a website. That's why I posted it here. I was hoping to get some help on it but apparently all I am going to get is snide remarks. I mean, why even reply?
 

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