The sum of the reciprocals of all primes diverges | #some2

It follows as a corollary that there are infinitely many primes, and proving this corollary is left as an exercise to the viewer.

I realized right after upload that I should have specified more clearly that lower case k stands for the number of small primes. Also, at 1:25 I done goofed (I forgot that my sqaures aren't unit squares in the software when I did the calculation). The real answer is closer to 10^(50 billion).

[hughcaldwell]

I saw this proof on Wikipedia a while back and absolutely loved it. Great presentation here.

[svaritjoshi]

Extremely elegant proof
I myself prefer a combinatorial or number theoretical proof over a calculus based one.
Also




WHAT WAS THE MOTIVATION BEHIND THIS PROOF !!!!
NEVER SEEN ANYTHING LIKE THIS BEFORE TO PROVE A DIVERGENCE!!

[SteveThePster]

It's a very clever proof, lots of random-looking functions and sums created to arrive at the contradiction

[doctorb]

Erdos visited the University of Florida a few times while I was a student. The Department always welcomed him .

[barakap]

At the end you can take 2^(2k + 2) out, so you have 2^(2k+2) * (1+2) = 2^(2k+2) * 3 at the right hand side, which is clearly less than 2^(2k+4) which is 4 * 2^(2k+2) thus a contradiction

[neelmaniar]

Super elegant and brilliantly explained! Made my day, thank you very much. I hope you keep making such videos!! I appreciate that you go through and respond to commenters, especially those that are critical or incorrect.

[baronhelmut]

This has pretty cool implications too. We know that the harmonic series diverges. And every time we decrease the exponent of the individual terms it converges. This should also be the case for the sum over the prime reciprocals. If we calculate the differences between the individual results of both sums up to the same reciprocal, we should be able to derive some sort of relation from it. But what kind of relation ? Nobody knows.

[lllULTIMATEMASTERlll]

The first time I watched this video I thought it was an ugly proof that was too complicated to be beautiful. But I watched it again and I think this proof slaps.

[JadeVanadiumResearch]

Very cool to see an elementary proof of this fact! Thanks to the algorithm for showing me this video randomly, which I seemed to have missed back during SoMe2 :)

If you know the Prime Number Theorem, there's a much faster (albeit less elementary) proof of divergence. Letting pi(n) denote the number of primes p<=n, the PNT states that pi(n)~n/ln(n). That is to say, the limit as n tends to infinity of pi(n)*ln(n)/n converges to 1. Letting p[n] denote the n'th prime, you can use PNT to prove that p[n]~n*ln(n). The limit is simple after substituting n=pi(k) and limiting k to infinity, which works since ln(pi(k))~ln(k) due to PNT, and p[pi(k)]=k by definition. When we are taking the sum of 1/p[n], we can rigorously approximate this with the summation over 1/(n*ln(n)) due to PNT and the previous arguments, and then deploy the integral test to prove divergence.

[harvek]

extremely elegant proof. honestly, very impressed by this result. also great animation!

[rosiefay]

Nice proof! At 2:23 you have a strict inequality $< 1/2$. So, at 6:11, you may take n=2^{2k+2}, which leads to 2^{2k+1} < 2^{2k+1}, contradiction.

[syllabusgames]

I like the presentation, especially when you copy an equation to the corner for later, so I can easily keep track of where things are coming from. I don’t understand prime factorization, so I got lost about 3 minutes in. I’m just not in the target audience. Otherwise, good video.

[disgruntledtoons]

I discovered this from how the zeta function approaches infinity as the argument approaches 1. The logarithm therefore approaches infinity. The logarithm of x is the sum of the logarithms of each factor of x. The factors of zeta(1) can be shown to be the sums of the negative powers of each prime. The logs of each such factor can be shown to be less than a corresponding value in the series equalling zeta(1). Therefore, the sum of the reciprocals of the primes approaches infinity.

[Alexander-ohry]

I found this very hard to follow compared to other proofs in video form. I think it is because the proof jumps around so much with seemingly unrelated ideas like prime factorization

[FineDesignVideos]

HigureSensei, I think the proof is not as random as you've presented it. The maths does come together very elegantly to give the conclusion and I think that elegance distracts us from asking our usual question: why?

As you mentioned in the video, the sum of reciprocals gives you a natural upper bound on the fraction of numbers that have one of these primes. With the definition of convergence, you can conclude that there's a tail of the infinitely many primes that can only account for at most half the numbers in any set [N].

Now for the finite set of primes before the tail, say there are k of them. These primes do generate infinitely many natural numbers, but it does so exponentially slowly. The pth number it generates is exponentially large in p. Specifically among the first N numbers, you can generate only (log N)^k many numbers.

So (log N)^k + N/2 should cover all of [N] but it can't when N is large.

I believe the whole square free part of Erdös proof was just to avoid a log and to keep the proof working with as elementary tools as possible (sacrificing understandability along the way). It is really cool that that choice does give such a neat explanation of what N is large enough.

[letismadeo]

Amazing! Such estimates always seem useless to me until I see the conclusion. For example when bounding the set of numbers with only small primes, the way the bound of n was incorporated was not something I'd think of

[StratosFair]

Very clear exposition of a very elegant proof, thank you !

[chadwick]

At 6:15, why can you set n to 2^(2k+4)? Isn't k uniquely determined by n since k is the number of primes required to represent the prime factorization of every integer up to n?

EDIT: I just saw the description. k only needs to represent the small primes. I think you'd still have to show that the number of small primes can grow slower than (log(n)-4)/2. But it's easy to see that if it grows faster than that, you end up with enough small primes to make the sum infinite anyway, so it makes sense.

[matemindak]

It feels super weird that this diverges but say 1/n^(1.01) converges

[nousernameleft]

i think you made the proof more complicated than the original version especially at the end 😅 but its a nice one