Does the alternating sum of prime reciprocals converge?

$$\sum_{n = 1}^\infty \frac{(-1)^n}{p_n}$$ where $p_n$ is the $n$th prime.

I have computed this to 10000 rather than infinity. My results suggest that convergence does happen but it's very slow. But I can't even be sure about the first few digits: $-0.26959$?

I have looked at the "questions that may already have your answer" and some of the "similar questions," but they involve somewhat different formulas.

• en.wikipedia.org/wiki/Alternating_series_test – Count Iblis Jun 18 '17 at 23:20
• More digits at OEIS A078437. – Raymond Manzoni Jun 18 '17 at 23:36
• How can it not converge? Are you expecting that it might shoot off to positive infinity? Or negative infinity? Or are you expecting it to e.g. oscillate without its amplitude going to zero? – Mehrdad Jun 19 '17 at 5:33
• @RobertSoupe: I see. Well, notice that it cannot possibly oscillate in the limit, because the terms approach 0. (Why? Because by definition, oscillation would mean the series forever goes up and down with some nonzero minimum amplitude. But the terms go to zero, so the amplitude gets arbitrarily small, i.e. it stops oscillating in the limit.) So the only thing it can possibly do is converge or diverge to infinity. (cont'd) – Mehrdad Jun 19 '17 at 13:13
• (cont'd) ...but it can't diverge to infinity, because every subsequent term goes in the opposite direction of the previous term with a strictly smaller step -- which means every sum up to a positive term is an upper bound on the entire sum, and every sum up to a negative term is a lower bound on the entire sum. So we have a finite upper and lower bound, meaning it can't shoot off to infinity. Given that it can't oscillate either, that means it must converge. (But note that I was just trying to convey the intuition here, not form a mathematical proof.) – Mehrdad Jun 19 '17 at 13:15

People already told you that you can prove this by the alternating series test; here's the intuition:

Notice that the series cannot possibly oscillate in the limit, because the terms approach 0.
(Why is this sufficient? Because by definition, oscillating in the limit would mean the series forever goes up and down with an amplitude that doesn't decay to zero. But the terms do go to zero, so the amplitude does get arbitrarily small, i.e. the sequence stops oscillating in the limit.)

So the only thing it can possibly do is converge or diverge to infinity.

But... it can't. Every subsequent term goes in the opposite direction of the previous term with a strictly smaller step—which means every sum up to a positive term is an upper bound on the entire sum. Similarly, every sum up to a negative term is a lower bound on the entire sum.
So we have a finite upper and lower bound, meaning the series can't shoot off to infinity.
Given that it can't oscillate in the limit either, that means the only thing it can do is converge.

(Note that this isn't a proof, e.g. I didn't even define "amplitude". This is only the intuition.)

• I don't agree to your last line. This is an informal but fully rigorous proof. Contrary to what many believe formalism is different from rigor and is more akin to "legalese of mathematics". +1 for an excellent answer. – Paramanand Singh Jul 2 '17 at 12:38
• @ParamanandSingh: Haha thanks, but I don't see why! I really do think I would really need to define "amplitude" before it becomes rigorous. The standard definition requires periodic functions, which we can't assume here. We also can't define it as the maximum absolute deviation from the limit, because that assumes the limit exists (which is what we're trying to prove), etc. – Mehrdad Jul 2 '17 at 13:12
• If you want the formal word for amplitude it is difference between limsup and liminf of a sequence. For a convergent sequence these are equal and thus amplitude is $0$. And the way we prove the convergence is Leibniz test is exactly by considering partial sum for even number of terms (these are increasing) and odd number of terms (these are decreasing) and they are obviously bounded. So both converge and because of zero amplitude they converge to the same value. – Paramanand Singh Jul 2 '17 at 13:53
• If you see the standard proof for Leibniz test, you will see that it echoes exactly the same argument as your answer but uses lot of math symbols instead of English words (like in your answer). That's why I said that your proof is rigorous. – Paramanand Singh Jul 2 '17 at 13:57
• @ParamanandSingh: Makes sense, thanks :-) – Mehrdad Jul 2 '17 at 14:04

Yes: any series of the form $\sum_n (-1)^n a_n$ with $a_n>a_{n+1}>0$ and $a_n \downarrow 0$ converges, by the alternating series test.

I'm sure you already know what the alternating series test is. I am answering more for the other people who happen onto this question.

Wikipedia is not completely wrong when it comes to math, but I still prefer to look for other sources. The link at the bottom of the page is to the Mathworld article on the Leibniz criterion.

Okay, it's a very short article that essentially says that if the sequence is "monotone decreasing," then the infinite sum converges. Then I looked "monotone decreasing:"

Always decreasing; never remaining constant or increasing. Also called strictly decreasing.

The sequence of prime numbers is the opposite of this, monotone increasing. But the alternating sequence you're summing is the reciprocals of the primes, and that's definitely a monotone decreasing sequence. Verify that this limit exists: $$\lim_{n \to \infty} \frac{1}{p_n} = 0,$$ which means we can get $\frac{1}{p_n}$ arbitrarily close to $0$ by choosing larger and larger primes.

Furthermore, as nearly everyone reading this ought to know very well, there are infinitely many primes. This sequence passes the alternating series test.

Of course that doesn't tell us how quickly convergence happens. In this case, it's very slow, and this number is not known to as many decimal places as some other infinite sums: $0.26960635197167$. According to Sloane's A078437, the next digit is believed to be $4$.

• BTW Wikipedia articles on math topics are much better than the mathworld pages. At least they convey some meaning whereas mathworld is just a list of cryptic formulas. Don't know what's the purpose of that. – Paramanand Singh Jul 2 '17 at 12:41
• They might look better but are not necessarily better. At least with Mathworld I know who to blame if I'm misinformed reading it. – Mr. Brooks Jul 5 '17 at 20:40