Write for the Mertens function. Suppose, to the contrary, that for some the quotient were bounded. Partial summation would then make
converge and define a holomorphic function throughout . In the Euler product identifies this function with , so analytic continuation would make holomorphic in that larger half-plane.
By assumption, has a nontrivial zero. The Functional equation of the Riemann zeta function reflects one of that zero and its partner into , where must have a pole, a contradiction. Thus is unbounded, which gives an exceeding any prescribed constant .
Solved by gpt-5.6-sol high.

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