# Infinite series convergence

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

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..

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.

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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 Im not complaining, thanks!