# Showing the alternating series $\sum_{n=1}^\infty (-1)^n \frac{n}{p_n}$ where $p_n$ is the $n$th prime converges

So what I want to prove is:

Proposition: Let $p_n$ be the $n$th prime. Then the alternating series $$\sum_{n=1}^\infty (-1)^n \dfrac{n}{p_n}$$ converges.

Here's my (original) attempt. Could someone verify my proof is alright?

Lemma 1: If $a_n$ and $b_n$ are sequences and $\lim_{x\to \infty} \frac{a_n}{b_n} = 1$, then

$$\sum_{n=1}^\infty a_n \text{ converges} \iff \sum_{n=1}^\infty b_n \text{ converges},$$ $$\sum_{n=1}^\infty a_n \text{ diverges} \iff \sum_{n=1}^\infty b_n \text{ diverges}.$$

Proof: Follows directly from the limit comparison test.

Proof of Proposition: Note that by the Prime Number Theorem, $$p_n \sim n \log(n),$$ that is, due to $$\lim_{n\to \infty} \frac{p_n}{n \log(n)} =1.$$

Thus, by Lemma 1, $$\sum_{n=1}^\infty (-1)^n \dfrac{1}{\log (n)} \text{ converges} \implies \sum_{n=1}^\infty (-1)^n \dfrac{n}{p_n} \text{ converges}.$$

Now, the series $\sum_{n=1}^\infty (-1)^n \dfrac{1}{\log (n)}$ converges (as made clear by the alternating series test). As a result, our proposition is proven.

Note (Clément C.): As mentioned in the comments, this particular argument is faulty, since the "Lemma" used does not hold. (Specifically, it only holds for positive sequences (or negative sequences), but not those whose sign alternate.) A proof of convergence (or divergence) of the original series would be quite interesting.

Observe also that the alternating series test does not seem to apply here, as even with the Prime Number Theorem it is not obvious (and may be false) that the sequence $\left(\frac{n}{p_n}\right)_n$ is non-increasing. Moreover, it is not clear that the "usual" remedy for this (i.e., performing a Taylor series expansion of $\lvert a_n\rvert$ to get a constant number of terms which constitute, each by itself, non-increasing sequences; until a last term is reached which is the term of an absolutely convergent series) can be applied here, as such a series development appears to only give very slowly decreasing terms. (That is, reaching a term whose seriess absolutely convergent does not seem to happen within a constant number of terms).

• I'm really shaky about where I applied the lemma. Feel things may have went sour around there. – Andrew Tawfeek Sep 14 '17 at 1:58
• Your "lemma," which does not have a proof, is false. The limit comparison test requires that all the terms be positive. Imagine that $a_{2n} =1/\log(n)$ and $a_{2n+1} = -1/\log(n)$ for $n > 1$. Then certainly $\sum a_n$ converges. But if $b_{2n} = 1/\log(n) + 1/n$ and $b_{2n+1} = -1/\log(n)$, then certainly $a_n/b_n \rightarrow 1$ but $\sum b_n$ does not converge. – user466572 Sep 14 '17 at 2:00
• @AndrewTawfeek Do you want to revive interest in the question, to get a correct answer? One way to do it would be to put a bounty on it. (I can take care of it if you don't want to spend reputation points.) – Clement C. Sep 16 '17 at 23:05
• Related (but with no answer to your question): math.stackexchange.com/questions/110009/… – Steve Kass Sep 17 '17 at 0:02
• in a side note : its easy to prove converges for $\sum \limits_{n=1}^{\infty} (-1)^n \frac{n}{\theta(p_n)}$, so it might give some evidence that $\sum \limits_{n=1}^{\infty} (-1)^n \frac{n}{p_n}$ could also converges. – Ahmad Sep 21 '17 at 16:39

It's an unsolved problem see Primes Sum Number 8

Because the gap between two consecutive primes $n^{0.53}\geq g_n\geq 2$ for large $n$ ,and by PNT the average gap is $\ln n$.

Now one can show that if most of the gaps are away from $\ln n$ then the sum diverges, but if most of the gaps are around $\ln n$ then the sum converge.

Even if used the bound given by the famous R.H. we still get that $\sqrt{n} \ln n \geq g_n \geq 2$ which does not help.

May be it could have a chance of being solved using Cramér's conjecture for the gap $O(\ln^2 n) \geq g_n \geq 2$ (i am not sure since its not solved).

The limit comparison test requires positive terms, as explained by user466572.

Possibly of more use: For $n \geq 6$, $$\log n + \log\log n - 1 < \frac{p_n}{n} < \log n + \log \log n \text{.}$$ (See Wikipedia: Prime Number Theorem:Approximations for the $n^\text{th}$ prime.)

But this is clearly not sufficient to conclude, the sequence not being decreasing. There is a plot of the partial sums of the series, showing why it could converge, but at such a slow rate that we'd probably need super-strong versions of the Cramer conjecture on the prime gap to show it • And ? ${}{}{}{}$ – reuns Sep 14 '17 at 2:53
• I don't know any way to study the convergence of $$\sum_n (-1)^n\frac{n}{p_n}=\sum_n\frac{n (p_{2n+1}-p_{2n}) -p_{2n}}{p_{2n}p_{2n+1}}$$ But the random model for the primes (that $p_{n+1}-p_n$ follows a geometric distribution of mean and standard deviation $\log n$) could imply it diverges. – reuns Sep 14 '17 at 3:03
• How do you conclude from what you wrote, @EricTowers? – Clement C. Sep 14 '17 at 3:18
• @ClementC. : That would be ... the alternating series test. – Eric Towers Sep 14 '17 at 4:35
• @reuns : Oh, well. It was a pleasant thought while it lasted... – Eric Towers Sep 14 '17 at 5:13