Understanding an elementary proof of prime number theorem in full details from scratch in six hours

This video series features the elementary proof of prime number theory (PNT) following Hardy's book "An Introduction to the Theory of Numbers" (in the spirit of Erdos and Selberg), targeting ambitious high school students, undergraduate students as well as graduate students who need to watch something to help them fall asleep faster at night. The audience are only assumed to know calculus, basic combinatorics (binomial theorem) and basic elementary number theory (prime numbers and prime decomposition). No prior knowledge of complex analysis (we won't need it) or arithmetic functions (will be introduced from scratch) is required. "Six hours" might be exaggerating to smart people since I believe you could play at 1.25x speed or even 1.5x speed. If you prefer to watch this piece by piece, a playlist of 20 short video sections can be found here:

[stanleydodds]

This was a great video. I've seen some complex analysis proofs of PNT, but I've never looked at this elementary proof, and this was a great walkthrough of it.

I was just going to make one point; at 5:42:47 we sum o(1) (as m goes to infinity) over all integers m from 0 to M-1 to get o(M) (as M goes to infinity). This is not entirely trivial, because the o(1) depend on each different integer in the sum going to infinity, but (for example) any specific term, or all of the "small" terms up to some constant, do not "go to infinity" in this way; the o(1) do not all go to 0 uniformly, you might say.

It's not difficult to prove that this doesn't matter though, by splitting the sum into the sum from 0 to sqrt(M), where there are only sqrt(M) bounded terms, so the sum is o(M), plus the sum from sqrt(M) to M, where each m > sqrt(M), so all go to infinity as M goes to infinity, so the o(1) hold uniformly and we get o(M). So maybe it was too trivial to be worth including. It's the same as the fact that the average of the first M terms of a sequence has the same limit as the original sequence, so it's a pretty standard result I suppose.

[flipflipshift]

Something I've thought off and on about is proving a slightly weak version via the following sketch:
1. Prove nth partial sum of reciprocal of primes is eventually more than than Clog(log(n)) for any C<1 (easy, something stronger has short proof on wikipedia)
2. Prove nth partial sum of reciprocal of primes is eventually less than than Clog(log(n)) for any C>1 (Euler provided a sketch in his argument that the partial sums diverged; need to formalize)
3. Show that nth partial sum of 1/klog(k) respects the same bounds (trivial)
4. Conclude lim sup p_n/(nlog(n)) >=1 and lim inf p_n/(nlog(n))<=1.

Any thoughts?

[samueldeandrade]

Oh wow ... man, the best math channels are hidden.

[omargaber]

Please can you explain the gaps between the prime numbers (Miner, Zhang)?
There is a paper about preparing a chemical that is also difficult and I did not understand it, which was done by this young man)Bertrand’s Postulate for Carmichael Numbers
Daniel Larsen). I wish that you explaine it.