Infinite series convergence

  • Thread starter camilus
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  • #1
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[tex]\sum^{\infty}_{x=1} \frac{cos(14.1347 \ln (x))}{x^{a}} = 0[/tex]

Is there a way to solve for a? I dont think so but maybe someone here will have an insight as to what to do..
 

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  • #2
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This equation is

[tex](1/2)\Re \zeta(a-ui) = 0 [/tex]

for [tex]u = 14.1347[/tex], where [tex]\Re[/tex] signifies the real part, and [tex]\zeta[/tex] is the Riemann zeta function.

The attached picture shows the graph (I did it without the factor 1/2). The .mw file is the Maple code that generated this picture.

So [tex]a=1/2[/tex] looks like the solution. If we replace [tex]u=14.1347[/tex] by the nearby zero of the zeta function [tex]u = 14.134725141734693790\cdots[/tex] then the solution would be exactly [tex]a=1/2[/tex] of course.

Caveat. Probably the original series converges only for [tex]a > 1[/tex] , so my analysis applies only to the analytic continuation.
 

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  • #3
146
0
This equation is

[tex](1/2)\Re \zeta(a-ui) = 0 [/tex]

for [tex]u = 14.1347[/tex], where [tex]\Re[/tex] signifies the real part, and [tex]\zeta[/tex] is the Riemann zeta function.

The attached picture shows the graph (I did it without the factor 1/2). The .mw file is the Maple code that generated this picture.

So [tex]a=1/2[/tex] looks like the solution. If we replace [tex]u=14.1347[/tex] by the nearby zero of the zeta function [tex]u = 14.134725141734693790\cdots[/tex] then the solution would be exactly [tex]a=1/2[/tex] of course.

Caveat. Probably the original series converges only for [tex]a > 1[/tex] , so my analysis applies only to the analytic continuation.

OMG brilliant! Thanks! I realized this a few days ago, and just realized now that I realized it, and it looks a bit clearer now. weird.. but Im not complaining, thanks!
 

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