Mathematician who solved prime-number riddle claims new breakthrough

Typically, your first step typically is to show that it is finite because that often is a lot easier, and if it turns out not being finite, you’ve spared yourself from doing some complex calculations.

The typical approach to show that some expression is finite is to replace difficult parts of the expression by simpler ones that certainly aren’t smaller. If the changed, definitely not smaller, expression is finite, so must be the original one.

As a simple example, a rouge estimate of sin(x) + cos(x) says both sin(x) and *cos(x) are in [-1,1], so their sum is in [-2,2]. That shows the expression is finite for all x. (Actually, the sum is in [-√2, √2], but if you don’t need to be that precise, why bother?)

Similarly, I wouldn’t know what the sum of i=1 to infinity of |sin(i)|/2^i is, but each term is positive and not greater than 1/2^i, so the sum isn’t greater than 1, and that series must converge.

The basic strategy of Zhang, as well as Goldston-Pintz-Yildirim, Maynard, and others is as follows: consider the numbers

n, n+2, n+4, n+6, n+8, ... n+70,000,000

and prove that "on average" at least two of them are prime. Prove this, and the result follows.

Now this is not literally true, the average spacing between primes is unbounded. The main technical work is in coming up with a weighting function, which is biased towards those n for which the "average" claim is likely to be true. Prove this, and you're done.

One then has to compute "the weighted average number of primes in n, n+2, ..., n+70,000,000", subject to the weighting function you chose. It's relatively easy to do in principle, but it requires various estimates for counting prime numbers, all of which come with error terms, and the hard part is to prove that the error terms don't accumulate enough to completely drown out the main term.

Work on this subject has proceeded along two lines: (1) come up with a more efficient weighting function, that does a better job of picking out the n we really want to count; (2) improving the error estimates in the various counts that come up. Both of these are intrinsically quantitative, and inevitably less precise than one expects to be true -- and the upshot is that if you aim for 2, you might prove 70,000,000 instead, which is still an amazing achievement.