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I'm pretty sure the imaginary part $\mathcal I$ of $f(n)=\exp\left(\frac{2\pi\cdot i}{\log_n(p_n\#)}\right)$ converges as $n\to\infty$, and probably to $0$.

I stumbled upon this result studying the Collatz conjecture and it seemed potentially interesting in respect of prime number theory. Is this a difficult result to prove and does it have any significance?

$f:\Bbb N\to \Bbb C$

$p_n\#$ is the $n^{th}$ primorial

Show that $\lim_{n\to\infty} \mathcal I(f(n))=0$

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    $\begingroup$ Note that $\log (x\#) = \vartheta(x)$. $\endgroup$ – Daniel Fischer Sep 2 '18 at 20:29
  • $\begingroup$ @DanielFischer looks like $\vartheta(x)$ is the Chebyshev function. Are you implying this fact leads to a proof, or just saying it may be worth looking at? I can get $f\to \exp\left(\dfrac{2\pi\cdot i\log(n)}{\vartheta(n\log(n))}\right)$ $\endgroup$ – samerivertwice Sep 2 '18 at 22:08
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    $\begingroup$ Chebyshev's bounds show that the argument of $\exp$ tends to $0$. For the question you asked in a comment under the answer, you need the PNT. (Yes, $n\sin \frac{2\pi}{\log_n (p_n\#)} \to 2\pi$.) $\endgroup$ – Daniel Fischer Sep 3 '18 at 12:08
  • $\begingroup$ @DanielFischer given your last comment I'm surprised to see the imaginary part of this function still falling below $2\pi$ here: desmos.com/calculator/4twapxqpz5 Am I doing something daft here? $\endgroup$ – samerivertwice Sep 12 '18 at 8:28
  • $\begingroup$ @DanielFischer although it appears at least locally to obey $\lim_{sup} n\sin \frac{2\pi}{\log_n (p_n\#)} <2\pi$ $\endgroup$ – samerivertwice Sep 12 '18 at 8:31
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First, we can rewrite $f(n)$ as

$$f(n)=\sin\left(\frac{2\pi \log n}{\log p_n \#}\right).$$

We can use the fact that $p_n\# =e^{(1+o(1))n\log n}$ to get

$$f(n)=\sin\left(\frac{2\pi}{(1+o(1))n}\right),$$

which, of course, goes to zero as $n$ goes to infinity.

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  • $\begingroup$ Thanks. I guess this is fairly obvious because $n$ increases without bound. I actually made an error in my question :( so will re-ask. $\endgroup$ – samerivertwice Sep 2 '18 at 22:26
  • $\begingroup$ Does it obviously/trivially follow that the imaginary part $\mathcal I$ of $f(n)=n\cdot\exp\left(\frac{2\pi\cdot i}{\log_n(p_n\#)}\right)$ converges? (Note the extra $n$) $\endgroup$ – samerivertwice Sep 2 '18 at 22:28

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