The sum of the reciprocals of all primes diverges |

[@hughcaldwell1034]

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

[@MasterHigure]

Thank you! I was planning on tackling the very similar proof of Bertrand's postulate next (whenever that may be), but Michael Penn just came out with a video on that, so maybe I'll hold off on that for now.

[@Fire_Axus]

YFWI

[@doctorb9264]

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

[@MasterHigure]

He was a traveller. Apparently, he would find mathematicians he thought worked on cool problems, and go live with them while they worked on it together, and then he went on to the next cool thing.

[@SteveThePster]

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

[@barakap2372]

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

[@johnchessant3012]

very elegant proof

[@MasterHigure]

Erdös has a knack for those.

[@Fire_Axus]

no

[@neelmaniar1454]

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.

[@baronhelmut2701]

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.

[@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

[@stighemmer]

Excellent prestentation!

[@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.

[@Alexander-oh8ry]

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

[@MasterHigure]

That's entirely fair, and yours is not the only comment with this complaint. It is my first video in this format, after all. It was never going to be perfect. If there are ever more, I'll see what I can do to keep it more coherent.

[@syllabusgames2681]

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.

[@MasterHigure]

It was a goal to make this as accessible as possible. But there are limits. In this case, yes, prime factorizations are kinda unavoidable.

[@matemindak384]

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

[@MasterHigure]

That's very closely related to Bertrand's postulate and all its cousins: For any number t larger than 1, there is a constant N such that there is always a prime between n and tn for n > N. So the primes are eventually at least as common as the powers of t. And since t can be arbitrarily close to 1, the primes are eventually SIGNIFICANTLY more common than the powers of t for any specific t, like 1.01. But yeah, it is mind blowing the first time you really think about it.

[@matemindak384]

@@MasterHigure Thats interesting, Ill look it up. Thanks

[@douglaszare1215]

@@MasterHigure Wouldn't that be related if @matemindak384 said 1/1.01^n rather than 1/n^1.01?

[@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.

[@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.

[@harvek4158]

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

[@letis2madeo995]

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

[@FineDesignVideos]

It's not necessary to do it that way. Bounding it by (log n)^k would work even better. I think Erdös wrote it that way so that it's even more elementary and logarithms don't appear in the proof.

[@chadwick3593]

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.

[@MasterHigure]

We fix a k just based on what we need "small" and "large" to mean, which is to say, the reciprocal sum of all the large primes must be at most 1/2. This is before we ever introduce n. k doesn't depend on n at all. Only after that, do we pick an arbitrary n, and use this division of primes into small and large to over-estimate the number of natural numbers less than or equal to n. And it then turns out, if n is picked to be large enough (the boundary of which does depend on k), this overestimate is an underestimate, and we have a contradiction.

[@chadwick3593]

@@MasterHigure Okay, I think I get it. I missed something really basic. If the sum is finite, then Ns has a maximum size, and k can just be a constant so it's independent of n_s. Thanks for explaining 2 years after the upload!

[@olbluelips]

Very cool

[@Holasiquetal]

What about the powers of inverses of primes?

[@MasterHigure]

If the exponent is less than one, each term gets larger, and thus clearly the series diverges. If the exponent is greater than 1, then we can easily compare to the series definition of the Riemann zeta function, and see that we have a subseries of a convergent series (with all terms positive), so it converges.

[@andrewwalker7276]

This is the prime zeta function, similar to how the harmonic series can be made into the Riemann zeta function. The wikipedia page en.wikipedia.org/wiki/Prime_zeta_function is well worth a read. along with the linked Carl Froberg paper if you can access it. It also has heaps of zeros which I've been calculating for maybe 20 years? They don't form a line!

[@rosiefay7283]

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.

[@EvergrowingInfinity]

Intriguing!

[@loicboucher-dubuc4563]

Phenomenal!

[@StratosFair]

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

[@Yougottacryforthis]

Very non standard calculus approach about a subject in the core of calculus - divergence. Kinda rad.

[@yuyan930]

this indicates that the primes are not very sparse

[@MasterHigure]

@@yuyan930 It does indeed. Speaking of sparsity, the primes is a cool example of another result and conjecture. It is known that any set of natural numbers with positive asymptotic density contains arithmetic progressions of arbitrary length. The primes have zero asymptotic density, but the sum of their reciprocals is infinite, which is a slightly weaker non-sparsity condition. The primes also contain arithmetic progressions of arbitrary length. It is unknown whether infinite-reciprocal-sum generally implies arbitrary-length arithmetic progressions.

[@rapid4188]

At 2:16, was that inequality sign meant to be a less-than? The wording used in the video like 'all the way *up* to L - 1/2' had me torn

[@wyboo2019]

now find the value of the alternating sum of reciprocals of primes

[@Karcrab_mcshouty]

and what about the sum of the reciprocals of all squared primes ?

[@MasterHigure]

@@Karcrab_mcshouty That's obviously less than the sum of the reciprocals of all the squares. I don't know what the result is, but it's finite.

[@CutleryChips]

Is the implication that primes grow faster than n but slower than n^2? Since sum of 1/n diverges but sum of 1/n^2 converges

[@MasterHigure]

@@CutleryChips And also slower than n^a for any a>1, yes. At least eventually it does.

[@nousernameleft999]

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

[@MasterHigure]

@@nousernameleft999 I might have. I honestly don't remember, as it has been a couple of years.

[@hassanalihusseini1717]

Nice proof!

[@martialversaux5746]

OK, this proof has gained popularity over the years and is visible in may vidéos. A more complicated challenge : what if we supress from th sum all prime(i) that contain a "1" ? Idem with a "2" ? The same until 9 ? Anybody interested with this amazing problem ?

[@MasterHigure]

@@martialversaux5746 If you do that just with the natural numbers, I think I heard somewhere that the sum converges (although I might be wrong here). If so, obviously the same is true for the primes.

[@martialversaux5746]

@@MasterHigure Thanks for this first remark. Clearly not simple to draw a parallel with the harmonic serie amputed from some digits...

[@giladperiglass5734]

Just saw this now, can anyone explain why I'm wrong? I'm only at the start of the proof, but I see that you're assuming that even if there was a limit L for the sum of all reciprocals of primes, you could sum the first K-1 reciprocals to get L-½. This isn't necessarily true, even if it was a convergent sum, since the sum could "skip around" L-½, if one reciprocal would get the sum to be slightly smaller, and the next to be slightly bigger than L-½.

[@MasterHigure]

I never say it should hit exactly (at least I don't mean to). Just that it should be *at least* L - 1/2.

[@fuxpremier]

When in doubt with limits, it's always useful to come back to the "epsilon-delta" (or with natural numbers, the "epsilon-N") definition of the limit. In this case, we just took epsilon = 1/2 and K is the corresponding N.

[@honeyinglune8957]

From my limited knowledge, this seems like a surprising result. Am i wrong?

[@MasterHigure]

@@honeyinglune8957 Not directly surprising, per se. But to someone who doesn't know any strong theorems about prime density, it's really a coin toss as to which way it goes. Although to me, many years ago, when I learned that the reciprocal sum diverges for the primes, but a series like the sum of 1/n^1.0001 converges, that certainly was somewhat a surprise.

[@znhait]

​@@MasterHigure Once you know that the numbers of primes are infinite, isn't it quite logical to say that it does in fact diverge. The reason is that since 1/n diverges, n being a natural number, for any given set, there are always more natural numbers than prime numbers. It follows that if p is a prime, 1/p is decreasing much slower.

[@MasterHigure]

@@znhait It is well-known (and not very difficult to prove) that the sum of 1/2^n converges. It is well-known (and a bit more difficult to prove) that the sum of 1/n^2 converges. There is absolutely no reason to automatically believe that the sum of 1/p diverges.

[@znhait]

@@MasterHigure I am not sure what your point is. I just made an observation that since n increases faster than p, there is reason to believe that 1/p has to diverge.

[@MasterHigure]

@@znhait No, it is, in fact, the opposite. 1/p diverging automatically implies 1/n diverges, but 1/n diverging yields exactly zero information on whether 1/p diverges.

[@SpaghettiToaster]

Is there a constructive proof for this? I wonder what that would look like.

[@MasterHigure]

@@SpaghettiToaster Much like the sum of the harmonic series up to 1/n is known to be equal to ln n (up to a bounded difference of about 0.577), a similar result is known for this series (it is equal to ln ln n, up to a bounded difference of about 0.26). So to make a sum of 1000, you would need somewhere close to e^(e^1000) terms. Is that the kind of result you're asking about?

[@SpaghettiToaster]

@@MasterHigure Well, that doesn't actually prove that the series diverges, or does it?

[@MasterHigure]

@@SpaghettiToaster It does. Because you can make ln (ln n) arbitrarily large. So you can make the partial sum of the series arbitrarily large. And what is the divergence of a series to infinity if not the ability to make arbitrarily large partial sums?

[@SpaghettiToaster]

@@MasterHigure You're right lol

[@LeoStaley]

Oof. And I thought the harmonic series diverged slowly

[@MasterHigure]

@@LeoStaley There is always a slower series.

[@jwjustjwgd]

I wonder if there is a slowest possible divergent series

[@MasterHigure]

@@jwjustjwgd There is *always* a slower series. For instance, you can take any divergent series you like, and halve all the terms. It now diverges slower. More powerfully, take any strictly increasing sequence of real numbers that diverge to infinity, and you can find a sequence of natural numbers such that their reciprocal sum diverges slower than that real sequence (greedy algorithm does the trick). And obviously, there is no "slowest diverging" sequence of real numbers. Because for one thing, you can just take any such sequence and take the logarithm of each term (and do something sensible about the finitely many negative terms you might get as a result, such as just removing them from the sequence). Or apply any other slowly-diverging-to-infinity function you like.

[@Redstoner34526]

I just thought of something big from this video

[@bigbigx2250]

You know it's interesting that'd be the case when the sum of natural numbers is -1/12

[@MasterHigure]

@@bigbigx2250 The sum of the natural numbers isn't -1/12. The sum diverges to infinity. There are other ways to associate a real number to a series, and many of those are so similar to the actual sum that we call them sums (e.g. Abel sums and Cesaro sums). And some of those methods associate the series of natural numbers with -1/12. But they aren't actual sums, and the sum of natural numbers is not finite.

[@anestismoutafidis4575]

Σ p prime 1/p= ♾️ Σ=1/(2×10^- ♾️ )= ♾️ p=(2×10^-♾️ )

[@YoutubSosetXui]

Didn't understand much either Im dumb or you skip steps in explanation

[@MasterHigure]

@@YoutubSosetXui Certainly I skip steps; it is basically impossible not to. If I am skipping too many steps for you, that doesn't at all mean you're dumb. Just that the subject matter is unfamiliar to you. Do not confuse lack of knowledge and experience with stupidity, those are completely different things.

[@Slinel]

smh this proof assumes that L>0.5

[@MasterHigure]

@@Slinel No. If L

[@pragyanpranay3681]

Wait, 1/2+ some stuff? It must be greater than 0.5?

[@MasterHigure]

@@pragyanpranay3681 If the potential prospect of L < 1/2 scares you, I guess it is possible to go through the trouble of proving that this doesn't happen.

[@Ricocossa1]

​​@MasterHigure Isn't it just trivial that L>1/2 since all primes are positive and the first prime is 2? Also it doesn't matter for the proof. You can choose any finite number smaller than L (and smaller than 1) instead of 1/2, and the conclusion will be the same.

[@MasterHigure]

@@jonasdaverio9369 I am pretty sure I never say the reciprocal of the large primes sum to exactly 1/2 (and if I do, that's wrong, as there is no reason we should be able to hit exactly 1/2 even if the series converges). I assumed the original comment here was made mostly in jest, and I have responded in this particular thread with that in mind.

[@numero7mojeangering]

Seems connected to the derivative of ln(x)

[@KaliFissure]

THE CASUAL PROOF Since we KNOW that the sum of squares equals 1. As soon as primes show they value any more, at any point, than the inverse squares, we know it diverges.

[@MasterHigure]

@@KaliFissure Unfortunately, that only proves that the sum is greater than π^2/6. The argument is not quite strong enough to prove that the sum is infinite.

[@sistajoseph]

I don't believe it. I can't prove it- I mean disprove it right now but it is obvious to me. I can't be the only person in the world who has some brains-I mean intuition. Here is the issue, I think we have been there before. Zeno of Elea had a similar problem, where he could not get that sum to terminate and concluded that motion was impossible, even though it was obvious that motion happens all the time. Even this writing is motion.

[@oanceatudor4443]

ehm... what?

[@kingoreo7050]

He was actually describing geometric series, which do converge (if the terms are always getting smaller). Primes spread out slower than any geometric series, so their inverses shrink slower, and divergence can happen

[@gabitheancient7664]

what are you even talking about do you think having vibes makes you disprove mathematical theorems?

[@asparkdeity8717]

bro is a professional waffler

[@Fire_Axus]

too hard to understand

[@literallydeadpool]

Same, some day I'll get to understand tho

[@FadkinsDiet]

What level of math did you get up to?

[@Tumbolisu]

Any notion of graphics or concrete examples got completely eradicated after 1:30. As such, I simply lost all interest. It was nothing more than a boring proof that you find in a paper or in a lecture. A great example of how the word "elegant" is very much subjective.

[@MasterHigure]

The word "elegant", as used in the context of mathematical proofs, by mathematicians, while subjective as anything else, does have objective criteria. The most important ones are being short, using more elementary machinery than one might expect, and using unexpected approaches. While my presentation might not be optimal, the proof I present in all is generality would definitely fall into that category for the vast majority of mathematicians. Erdös had a knack for that. Not to say your dislike is invalid, I'm just being defensive and pedantic about that specific word. It is true that this video suffers from being mostly inspired by how I would present the proof to myself on a blackboard. I am sorry I couldn't cater to your tastes this time. Maybe I can learn to do better.

[@DarinBrownSJDCMath]

This is a very famous proof admired by mathematicians. It even appears in a book "Proofs from The Book", which says it has "compelling beauty".

[@Tsbwi82]

What a shame we arent doing concrete example but general proof

[@aureliensablon5428]

Quite a bad and just random bullshit go proof. The best one use just a bit of uniform continuity and summation. Here's the main idea. Each number can be classes by it's number of prime factor. As such by writing Phi(s) =PrimeSum( p^(-s)) we directly gets Zeta(s)=Sum( Phi(s)^n)=1/(1-Phi(s)) where it makes sense. If Phi(1) converge then Phi is continuous on [1,+infty[ or Zeta converge for all s>1 as such L(s)1 (because take the 3 first terms). This proof just use both arithmetical and topological tool necesseray to understand the basic non holomorphic properties of Zeta. That's why I find it perfect.

[@MasterHigure]

I happen to like proofs that use random elements that seem to come out of nowhere, but somehow still work. This is, of course, personal preference. The fact that you do not like these kinds of proofs does not make them bad and bullshit. It just means you don't like them.

[@gabitheancient7664]

@@MasterHigure tbh I don't think this is actually so random studying for math olympiad nt I started to get really used at estimating sets and stuff and coming to conclusions through this for example and I can see how someone would have the idea of separating the concept of "large" and "small" primes in that way to see if they can come to some conclusion, and testing a lot on what L- ? should be

[@rosiefay7283]

But the proof presented here doesn't use such advanced concepts as phi or zeta or holomorphic.

[@aureliensablon5428]

@@rosiefay7283 you just need that continuous function are stable under uniform convergence.

[@asparkdeity8717]

anyone can understand this video's proof, hardly anybody not studying math can understand yours. Yours is far more "bullshit" for the wider audience