Can the value of a be solved for in the infinite series convergence equation?

[camilus]

\sum^{\infty}_{x=1} \frac{cos(14.1347 \ln (x))}{x^{a}} = 0

Is there a way to solve for a? I don't think so but maybe someone here will have an insight as to what to do..

 

[g_edgar]

This equation is

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

for u = 14.1347, where \Re signifies the real part, and \zeta 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 a=1/2 looks like the solution. If we replace u=14.1347 by the nearby zero of the zeta function u = 14.134725141734693790\cdots then the solution would be exactly a=1/2 of course.

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

 

[camilus]

This equation is

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

for u = 14.1347, where \Re signifies the real part, and \zeta 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 a=1/2 looks like the solution. If we replace u=14.1347 by the nearby zero of the zeta function u = 14.134725141734693790\cdots then the solution would be exactly a=1/2 of course.

Caveat. Probably the original series converges only for a > 1 , 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 I am not complaining, thanks!