Without some additional structure, there shouldn’t be any “free” way of handling the intermediary region if the sign structure is random. I think we can recognize a limit of the method to understanding $latex \sum_{X < n < X + X/Y} a(n) v(n/X)$ for some general coefficients $latex a(n)$ in the following weak heuristic: it may be that the "bumps" of the sizes of $latex a(n)$ perfectly align with the signs and bumps of $latex v(n/X)$. We typically assume that $latex v(t)$ decays smoothly and monotonically from $latex 1$ to $latex 0$ on the interval $latex [1, 1 + 1/Y]$, but in fact the properties we use are indistinguishable from *any* smooth function on that interval that is $latex 1$ and $latex 0$ at the left and right end-points, respectively. So we are not using any distinguishing property of $latex v$, and so we cannot hope to account for perfectly aligned phases.

Relatedly, it is very annoying to get first moments in general. But if we can understand $latex \sum \lvert a(n) \rvert$, then of course we're fine just by bounding in absolute values. But we don't have a generically good way to get at this absolute value.

Fortunately, we sometimes (rarely) get a bit lucky. If we wanted to understand $latex sum a(n)^2$, the (non-normsquare) $latex \text{GL}(2)$ Rankin-Selberg coefficients, we can bound the intermediate region by referring to $latex \sum \lvert a(n) \rvert^2$ still, as we have access to this object. More generally, we can understand $latex \sum a(n) b(n)$ (coming from two different $latex \text{GL}(2)$ objects), since we can Cauchy-Schwarz-Buniakowsky and handle the individual sums through their Rankin-Selberg $latex L$-functions.

But considering $latex \sum a(n)$ where $latex a(n)$ comes from a totally generic "well-behaved" Dirichlet series, is totally hopeless (I think).

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