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No one knows if this sum $$\sum_{n>0}\frac{(-1)^nn}{p_n}$$converges (where $p_n$ is the $n$th prime). But what makes proving this so hard? We know that $$\bigg|\sum_{n>0}\frac{(-1)^n}{p_n}\bigg|<\frac{1}{2}$$converges and is similar to the other sum.

Remark: Apparently: $$\sum_{n>2}\frac{(-1)^n}{\pi(n)}$$converges, where $\pi(n)$ is the prime counting function.

Kamal Saleh
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    Where did you see the claim this isn't known? The summands are essentially $\frac{(-1)^n}{\log n} $ so should converge by the alternating series test. In fact, we have $p_n > n (\log n + \log \log n - 1)$ with the RHS monotonic so a comparison test plus alternating series test should suffice, no? – Brevan Ellefsen Dec 16 '22 at 05:04
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    @BrevanEllefsen https://math.stackexchange.com/questions/20555/are-there-any-series-whose-convergence-is-unknown – Kamal Saleh Dec 16 '22 at 05:10
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    @BrevanEllefsen It is true that $p_n\sim n\ln n$, however the terms are monotonic iff $n(p_{n+1}-p_n)>p_n$. This fails infinitely often if, for example, the twin prime conjecture (or much weaker conjectuers) are true, so the alternating series test cannot be applied directly. How exactly do you intend to use the comparison test when the comparisons are opposite for even and odd parity? – anon Dec 16 '22 at 05:15
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    For illustration, it is possible to have $0<a_n<b_n$ for all $n$ with $b_n$ monotonic decreasing to $0$ and yet $\sum(-1)^na_n$ diverges. This is because it is unknown if the differences $a_n-b_n$ (or in OP's case, the "error" terms for an asymptotic $a_n\sim f(n),$) for even and odd $n$ "cancel" each other out enough. A quick example is $a_n=(2+(-1)^n)/n$ and $b_n=4/n$. – anon Dec 16 '22 at 05:20
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    There is a simple theorem which applies to all the cases we know - an alternating decreasing function to $0$ will yield a convergent sequence. $n/p_n$ does go to $0,$ but we don't know it is decreasing. – Thomas Andrews Dec 16 '22 at 05:22
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    @ThomasAndrews You appear to be implying that the only cases of alternating series $\sum_n (-1)^n x_n$, $x_n > 0$, that we know converge are where $x_n$ are decreasing to $0$. For appropriate values of "we", that's not true. But it is true that the lack of such decrease prevents an easy proof of convergence in the case at hand. – Robert Israel Dec 16 '22 at 05:36
  • Just to note: The second series converges by the alternating series test, so its convergence could be determined by a [reasonably bright] calculus student. I assume you have a typo on the third series (it currently contains an undefined term), but if you correct it to be over $n\geq 2$, then its convergence and value are also easy to find, coming only from the fact that all but the first prime gap is even, giving the sum the value of $1$. [or perhaps you meant $\sum_{n > 0} (-1)^n / \pi(2n)$, which would be less trivial?] – Brian Moehring Dec 16 '22 at 05:40
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    My point with the previous comment being that you have a series that doesn't satisfy the assumptions of Leibniz's alternating series test, one that does satisfy those assumptions, and another one [after the presumed fix] that telescopes. Any perceived similarity between any two is purely cosmetic. – Brian Moehring Dec 16 '22 at 05:48
  • @BrianMoehring We could just compare the sum with the alternating reciprocals one (which is why I put $\ln 2$) – Kamal Saleh Dec 16 '22 at 15:18
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    To the well intentioned downvoter, why did you decide to do this action? – Kamal Saleh Dec 16 '22 at 16:09
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    I actually like the question and upvoted, but you gotta be careful with $\sum_{n>0}\frac{(-1)^n}{p_n}<\ln 2$ because as it stands the sum is negative so being less than $\log 2$ is irrelevant, while if you use absolute value or simply use $\sum_{n>0}\frac{(-1)^{n+1}}{p_n}$, the correct way to bound it is to truncate it before a negative term so, you get bounds $1/2, 1/2-1/3+1/5, 1/2-1/3+1/5-1/7+1/11$ etc, while you get inferior bounds by truncating it before a positive term so the sum is at least $1/2-1/3, 1/2-1/3+1/5-1/7$ etc – Conrad Dec 16 '22 at 18:21
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    The general shape of this question is "Why is it hard to prove that $\sum na_n$ converges, given that we know $\sum a_n$ converges?" To me, the answer is that we shouldn not expect knowledge of $\sum a_n$ (with its smaller terms) to help us understand $\sum na_n$ (with its larger terms). – Greg Martin Dec 16 '22 at 18:35
  • @BrianMoehring I will edit, thank you – Kamal Saleh Dec 16 '22 at 19:58
  • @runway44 This would make a good answer. – Jair Taylor Dec 16 '22 at 21:12
  • Is there any well-known conjecture that would imply convergence or divergence of this series? – Jair Taylor Dec 16 '22 at 21:16

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