Suppose that $z$ is some complex value. Is it possible to prove that

$$\lim_{n \rightarrow \infty} \sum_{j = 1}^n {\sqrt{n \over j}} \cdot \cosh(z \log {n \over j}-\operatorname{ Arccoth} (2z)) $$

cannot converge if $\Re(z) <> 0$?

Empirical testing strongly suggests both that it doesn't converge, and that the bounds of both the real and imaginary parts of the limit grow without bound as n increases, but I'd be really curious if there's a way to prove this.

  • $\begingroup$ If $\textrm{Re}\, z>0$ (say), then you're essentially summing $(n/j)^{1/2+z}$, so you would need $\sum_{j\le n} j^{-1/2-z}\lesssim |n^{-1/2-z}|$, which should not be hard to rule out. $\endgroup$ – Christian Remling Dec 13 '15 at 18:04

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