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From this short Mathematica program while investigating the convergence of the Dirichlet series for the Möbius function:

Clear[a, b, s, x];
s = 1/2 + 100*I;
Limit[1 - Sum[1/a^s, {a, 2, x}] + 
  Sum[Sum[1/(a*b)^s, {a, 2, x}], {b, 2, x}], x -> Infinity]

I get the output:

enter image description here

where Out[186]= is left unevaluated.

Does that mean that the limit exists but it is not known in the form of standard functions?

Mats Granvik
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    Usually unevaluated means that it doesn't have any rule that it can use to transform something into something else, otherwise it would have used that rule and then tried to see if there are even more rules that it knows. –  Mar 09 '21 at 11:53
  • SumConvergence[1/a^s, a] produces False and Sum[Sum[1/(a*b)^s, {a, 2, Infinity}], {b, 2, Infinity}] produces a warning about the divergence. – user64494 Mar 09 '21 at 11:54
  • Is it then also possible that the result is inconclusive? – Mats Granvik Mar 09 '21 at 11:55
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    For x=100, x=300,x=400 the command N[1 - Sum[1/a^s, {a, 2, x}] + Sum[Sum[1/(a*b)^s, {a, 2, x}], {b, 2, x}], 20] results in 2.3467599530378114392 - 0.2374475269288298657 I,1.8074308438887463892 + 0.0811844979002236090 I, and 2.4970344423667568855 - 0.4388452308824189010 I. These results suggest the limit under consideration does not exist. – user64494 Mar 09 '21 at 12:07
  • A ListPlot must always be suspect, but s=1/2+100*I; t=Table[1-Sum[1/a^s,{a,2,x}]+ Sum[Sum[1/(a*b)^s,{a,2,x}],{b,2,x}],{x,1,200}]//N; ListPlot[{Re[t],Im[t]}] gives me a hint. Yes, I also thought that was a nice plot, but I tend to like visuals. –  Mar 09 '21 at 12:44
  • That is a nice ListPlot. – Mats Granvik Mar 09 '21 at 12:45
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    From the help of Limit: Limit returns unevaluated or an Interval when no limit can be found. – Daniel Huber Mar 09 '21 at 13:32
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    Ask it at MSE.This is rather math than Mathematica. – user64494 Mar 09 '21 at 15:01

1 Answers1

3

First, we consider

 s = 1/2 + 100*I; Sum[1/a^s, {a, 2, x}]

-1 + HarmonicNumber[x, 1/2 + 100 I]

Second (The crucial tool is an assumption.),

Sum[Sum[1/(a*b)^s, {a, 2, x},Assumptions -> b \[Element] PositiveIntegers], {b, 2, x}]

(-1 + HarmonicNumber[x, 1/2 + 100 I])^2

At last,

Limit[1 + (-1 + HarmonicNumber[x, 1/2 + 100 I])^2 - (-1 + 
HarmonicNumber[x, 1/2 + 100 I]), x -> Infinity]

ComplexInfinity

Addition. The OP additionally requires a*b<=x. In fact, this is not any restriction because the limit as x->Infinity is taken. The sum of a double series does not depend on an exhaustion. Since

ComplexExpand[1/Abs[(a*b)^(\[Sigma] + I*t)]]

(Sqrt[a^2] Sqrt[b^2])^-\[Sigma] E^(t Arg[a b])

, the convergence depends only on \[Sigma]. It's more or less clear, that \[Sigma] should be greater than 1 to this end (I don't go into math deep.). Therefore,

Sum[Sum[1/(a*b)^\[Sigma], {a, 2, Infinity}, Assumptions -> \[Sigma] > 1 &&
 b \[Element] PositiveIntegers], {b, 2, Infinity}, Assumptions -> \[Sigma] > 1]

1 - 2 Zeta[\[Sigma]] + Zeta[\[Sigma]]^2

The same with Sum[1/a^s, {a, 2, x}]. Making use of Mathematica, we obtain the same answer as in MSE. The true convergence of a series is its absolute convergence. In view of it the series from the question are not absolutely convergent for s = 1/2 + 100*I.

user64494
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