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Are there results, or can someone point me to reading, on computations of the following sort:

$\lim_n \sum_{2 \le x\le n} (-1)^ { \log \log x }$

It would seem that there is reason to believe this should be something like $-\log(n)$. For example, see the graphic on this question which is the visual representation of $\sum_x (-1)^{\omega(x)}$ which for large $x$ is like $\sum_x (-1)^{\log \log (x)} = \sum_x e^{i \cdot \pi\cdot \log \log x }$. From the picture expectation, this looks something like $- \log n$.

To start, we note that we can reduce

$\sum_{2 \le x\le n} e^{i \cdot \pi\cdot \log \log x } = \sum_{2 \le x\le n} \log(x)^{i \cdot \pi} $.

Now this looks like it's related to a "more general zeta function" $\zeta (\log (n), -i\pi)$ (As opposed to the "standard" $\zeta(n,-i\pi)$.

From here, it would seem that the problem is likely out of reach from basically a college undergrad math student. Any textbooks/links for approaching this (hopefully that can ease me into it) would be nice, thanks.

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    $\begingroup$ Presumably 'cis' means $\exp(i\cdot) : t \mapsto \cos(t) + i\sin(t)$, so why not say that? Also, does $\lim_n \sum_n$ mean $\sum_{n = 2}^\infty$? $\endgroup$
    – LSpice
    Commented Jul 25, 2019 at 20:31
  • $\begingroup$ Sure, I edited the question to make it more clear $\endgroup$
    – Raj Raina
    Commented Jul 25, 2019 at 21:08
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    $\begingroup$ The limit will not exist. $\log \log n$ grows so slowly that the sum will be overwhelmingly close to $n(-1)^{\log\log n}$. $\endgroup$ Commented Jul 25, 2019 at 23:43
  • $\begingroup$ Actually, log log x - 1 is log log (x^(1/e)), so there will be some small cancellation/deviation. Say the sum is more like (n-n^{1/e})(-1)^{log log (n/2)), or something similar. Gerhard "There With A Delicate Touch" Paseman, 2019.07.25. $\endgroup$ Commented Jul 25, 2019 at 23:48
  • $\begingroup$ Well, at least $\int_3^\infty \exp(i\pi\log\log x)\;dx$ diverges. $\endgroup$ Commented Jul 26, 2019 at 11:22

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