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.
 A: 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 $4$.
A: 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.
A: 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.)
