One line proof by contradiction using probabilities: If the sum of 1/p_i converged, the value of the sum would have been a pretty famous constant thus I probably would have heard of it by now
But you neglect the case where it converges to an already famous constant. Nobody would take note if it was just another formula that evaluates to some multiple of pi, for example.
@@DevinBaillie true, therefore one needs to calculate the terms of the series until it exceeds 7, which would conclude since there are no famous constants bigger then 7
The proof that the sum at 7:47 diverges can be made simpler by realizing that 1/(1+N) > 1/2N for any N>2 so the terms of the sum can be compared 1:1 with harmonic series divided by constant 2*P1*...*PN. And since infinity divided by any finite number is still infinity, the series must diverge. No limit necessary.
@@mushyomens6885 The proof by @kasuha is valid and a bit simpler than the one in the video. The harmonic series is Σ 1/N which is known to diverge. We can always factor out a constant from an infinite sum so Σ 1/(2*N) = 1/2*Σ1/N is 1/2 *( a divergent series) which, of course, also diverges. I hope this has explained the problem for you.👍
Very neat! Euler's original proof for this is also pretty cool, he recalls the factorization of the harmonic series, namely the product over primes of 1/(1 - 1/p), and takes the log of both sides to get that the sum over primes of log(1/(1 - 1/p)) diverges. He then Taylor expands this to get the sum over primes of 1/p + 1/(2p^2) + 1/(3p^3) + ..., where the sums of 1/(2p^2), 1/(3p^3), etc. are subseries of 1/(2n^2), 1/(3n^3), ..., which converge. Thus for this to diverge it must be the case that the sum of 1/p diverges.
Though each sum of 1/np^n converges when n is larger than 1, the infinite sum of these sums does not necessarily converge (it does but this needs to be proven rather than just stated)
This proof is quite elementary and easy compared to other proofs I have seen. I had pondered this problem but never find an answer myself, this proof is the most satisfying one.
Agreed, it is comparable to the proof provided by Erdös. This one requires minimal facts about primes and more series manipulation, while the Erdös proof has minimal series manipulation but translates the problem to a counting one.
It seems to be by James A. Clarkson from 1966. Wikipedia has a link to it. en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Nice proof. Heard of similar proofs, but never this one. Just wanted to add that it is very "natural" to try to move from a problem with prime numbers to a problem with integers, since (1) the integers are generated by the primes, and (2) it has a much better structure which we can use (it has addition and multiplication) that the primes don't have. The next step would be to move from the integers to the real line, which has an even better structure - we have continuity, differentials, integrals, and all of the rest of the tools from analysis. For example, this is usually how you show that the Harmonic series diverges, by comparing it to the integral of 1/x. We can also do it on the other direction, for example trying to show that the sum of reciprocals of primes which are 1 mod 4 diverge to infinity, by moving to the same problem with all of the primes. In particular, this leads to one of Dirichlet's theorems that shows that there are infinitely many primes which are m mod n whenever gcd(m,n)=1.
According to the Müntz-Szász theorem, this implies that every continuous function on [0, 1] is the uniform limit of a sequence of polynomials where all the exponents in all the polynomials are prime numbers.
There was a man long ago who did work on the reciprocals of primes. He saw after how long the decimal digits of the reciprocals of primes repeated and he calculated them upto many thousands of primes. When his book was inspected, you see some corrections he made and they are either halving or doubling the original number that he had entered. He clearly had some formula for calculating them because if they were counting mistakes the corrections would be off by 1 or 2 or some other small number but he made corrections that showed he knew some sort of formula for the number of digits after which the decimal digits of the reciprocals of primes repeat. I saw this in a numberphile video and was curious. Thus video reminded me of that. Does anyone know more information about this?
I remember that Numberphile vid. Here's the rough idea: Let's set 1/p = 0.[n][n][n]... where [n] is some repeating string of digits. And let's say [n] is A digits long. That means we can multiply by 10^A and get: (10^A)/p = [n].[n][n][n]... = [n] + (1/p) Multiply LHS and extreme RHS by p: 10^A = p[n] + 1 Remember, [n] is just some integer. So the RHS is just one greater than a multiple of p. So we can reduce mod p: 10^A =(congruent) 1 (mod p) Now, Fermat's Little Theorem tells us that 10^(p-1) =(congruent) 1 (mod p), so long as gcd(10,p)=1 (which it will be for p =/= 2 or 5). From this it can be quickly seen that A must divide (p-1). You can also just say the same thing using Lagrange's Theorem from group theory instead. (I can expand on the Fermat/Lagrange step if you want, this is just the summary of the argument) Anyway, once you know that A must divide (p-1) you have much fewer numbers to check, and there are number theory shortcuts that can make those calculations (mod p) faster. And presumably this explains how he was out by a factor of two sometimes - 2 will always be a factor of (p-1) for the primes we're interested in, and (10^A)^2 = 10^(2A) (or vice versa) hence easy tog et a factor of two, remembering that (+/-1)^2 = 1.
My favorite proof of the divergence of this series utilizes the prime number theorem. The prime number theorem states that the prime counting function π satisfies the property that lim π(m)/(m/ln(m)) (m -> ∞) = 1. What this implies is that the mth prime number p(m) satisfies lim p(m)/(m·ln(m)) (m -> ∞) = 1. Therefore, the sequence of partial sums of 1/p(m) is asymptotic to the sequence of partial sums of 1/(m·ln(m)). Now, we can use the integral comparison test, noting that that the integral on [e, x] of 1/(t·log(t)) is bounded from above by the sequence of partial sums of 1/(m·ln(m)). But, hey, wait a minute! The integral can actually be evaluated explicitly, and it is equal to ln(ln(x)). But, look, lim ln(ln(x)) (x -> ∞) = ∞ !! Therefore, the integral diverges, so the sequence of partial sums of 1/(m·ln(m)) diverges, so the series of reciprocal prime numbers diverges. Q. E. D.
@2:29: Should be: If |x| < 1, then geometric series converges. Counterexample to video's statement: If x = -2, then x < 1, but the geometric series does not converge.
9:00 here's a more rigorous proof: Set p1 * p2 * .... * pn = k 1 / (1 + ik) > 1 / (k + ik) = (1/k)(1/ (1 + i)) Since k > 1 most definitely, we made the denominator bigger so the entire term is smaller Sum i from 1 to inf of 1 / (1 + ik) > Sum i from 1 to inf of (1/k)(1 / (1+i)) = (1/k)(Sum i from 1 to inf of 1 / (1+i)) = (1/k)(Sum of harmonic series - 1) = ♾️
I made a video covering Erdös' proof of this claim a couple of months ago, and I thought that proof was elegant enough (he likewise splits the series into two, but then uses that to show that there are fewer than N natural numbers in {1, 2, ..., N} for some large N, which is a contradiction). Why have I not seen this one before? It's wonderful! Where / when / who is it from? Little side note: You don't need contradiction here. You've really just shown that for any N, we get X>1. There is no actual need to ever assume X
Nice proof! This made me think about how this series converges even more slowly than the harmonic series and wether there's a way to construct the slowest converging series possible or maybe an iterative method for log(log(log...(N)...)) converging series
I had thought about this some while ago, and concluded that it wasn't possible. Suppose there was a sequence u_n such that : - u_n is positive and increasing - u_n diverges (u_n -> + inf) - for all positive divergent sequence v_n (v_n -> +inf), there is a > 0 such that for all n : v_n / u_n > a. For simplicity, I will use the shorthand u_x for a positive real x defined as : u_x = (x - Floor(x)) u_Floor(x) + (Floor(x + 1) -x) u_Floor(x+1). Consider now the sequence w_n = u_(u_n). We have Lim [n -> +inf] w_n / u_n = Lim [x -> +inf] u_x / x. Using the third point with v_n = log(n), we have a positive a>0, st u_n x] (1/x) * Product [k : 1 -> K] 1/log^k(u) = log^(K+1) (x) For example : Int [u : 1 -> x] 1/u log(u) log(log(u)) = log(log(log(x))) Considering the series z_n = Sum [p : 1 -> n] (1/p) Product [k : 1 -> K] 1/log^k(p), A series integral comparison would probably conclude that z_n is equivalent to log^(K+1)(n).
Thank you! I knew that this series converges but never saw a proof. A suggestion: Can you make a simular video (for us non-mathematicans) about the series of the rezipricals of the sum of the twin primes?
Thank you Dr Bazett , I really enjoyed this discussion and proof. It's interesting to think about series that are on he borderline of convergence/divergence. It seems intuitive that the reciprocal of primes might converge, because primes are sparse compared to integers for large numbers .
Second time through it made sense. So to prove the series diverges you throw away the early terms, create a new series then show that if you throw away enough terms from that you get a series that you recognise as divergent. Yes, that makes sense.
In the beginning I was kinda hoping the series would converge. Would've been cool if there was a number that's the sum of the reciprocals of all the primes. But I don't want to break reality, so this is fine, too.
this suggests a rather funny "proof": if the series converged to some constant, that constant would be famous enough that I would've heard of it. therefore it must diverge :)
1 / ( 1 - r ) for 0 < r < 1 will be greater than one for r = 1/2 1/2 + 1/4 + 1/8 ... is not greater than one let sigma for geometric series shown at 2m34s commence with i=0, or let sum be ( 1 / ( 1 - r ) ) - 1 ... superfluous outer parentheses added for emphasis
I guess it would have been possible to consider the series of 1/(-1+ i* p_1...p_N) instead which is greater than 1/(p_1...p_N) * harmonic series, hence diverges without comparison argument.
This can be done a bit more powerfully to show that there are infinitely many primes. When you use the fact that 1/(1+p1p2..pn) has only prime factors above pn, you are implicitly using the fundamental theorem of arithmetic which is proven using the fact there are infinite primes.
I don't think that uses the fundamental theorem of arithmetic. In its simplest form, unique factorization means "p | ab => p | a OR p | b" for all irreducible numbers p and integers a,b, and this we never use. If you factorize (not necessarily uniquely) A = 1 + k * p_1 * ... * p_N into irreducible numbers A = q_1 * ... * q_M, then it is clear that none of the q_j can be a p_i, since A mod q_j = 0 for all j and A mod p_i = 1 for all i. So any number A_k of this form is the product of irreducibles q_j > p_N (perhaps in more than one way), which means 1/A occurs in X+X^2+X^3+... (perhaps multiple times). So X+X^2+... >= sum 1/A_k = inf.
For those of us who aren't calculus professors there is no need for any" limit comparison test". Otherwise the proof is extremely beautiful and simple!!!!!! But this doesn't mean is not inventive! Thank you a lot for this new proof.
So, say that you have a sequence consisting of the sums of the first n terms of a sequence like (1/2,1/(2*3),(1/(2*3*5),...,1/p_n#) and then subtract the results from 1. So like (1/2, 1-1/2-1/(2*3), 1-1/2-1/(2*3)-1/(2*3*5),...,1-1/2-1/(2*3)-1/(2*3*5)-...-1/(p_n#)). This sequence converges to 0 per the Prime Number Theorem. Is there any way to describe the behavior of this sequence at finite positions (as opposed to at n=infinity)?
Very nice video. My only suggestion is to mention we are dealing with series which monotonically converge. This helps with notions of convergence of subsequences.
What about the sum of reciprocal of the square of primes- it definitely converges because it's smaller than Basel problem which we know it's equal to (pi^2)/6. It will be interesting to see to what it converges.
I would just use direct comparison as 1/(1 + i × prime product) > 1/(2 × i × prime product) = 1/2 × 1/(prime products) × 1/i Also Sum 1/i diverages is just showing its bigger than 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 +.... Which will diverage to Infinity. Thus this can be shown with these arguements to College Algebra or advanced Int Alg (High Schoolers) students without any calculus ideas.
Your end claim appears to work, but only because both series contain only positive numbers- it’s relatively easy to imagine a convergent series containing a divergent subseries if the convergent series contains a term (-1)^i (or sin(i) if we feel fancy) A well known example would be sum i = 1 to inf {((-1)^i-1)/i = ln2 We can recognize the shifted geometric series 1+1/3+1/5+… is a subset of the ln(2) summation, diverges to positive infinity, and is at all points greater than the sum of prime reciprocals, yet the parent summation converges. I would need further consideration on whether or not similar slight of hand could be used for a series of positive values can converge as a subseries diverges, and my instinct is to say it cannot (an infinite positive term will always overwhelm a finite term in the limit) but I’d be curious if mathematicians in comments can prove my instinct wrong
I really enjoyed this video. I almost passed it by because of the “shocked” click bait thumbnail. It was truly the title that convinced me to watch the start, and I was pleasantly surprised to see that you’ve got a semi-Mathlogger style for a topic I’d not thought about before. But normally I skip the shocked pikachu thumbnails. I haven’t subbed yet.
Ha. To be honest, I find them a bit cringe myself. But the analytics is overwhelmingly clear that when I give those types of thumbnails the click though rate is just higher. It's a wierd phenomenon
@@DrTrefor that’s fair. I get that the algorithm can encourage creators to encourage creators to “work it”. I really loved your video. I’ve subbed, since I’d rather see more of your style than less. I hope you get big enough that you don’t need to shock me every time!! 😂
Summary: Assume sum(1/p_n,n=1..) converges. Then for some N, X = sum(1/p_n,n=N..) < 1. Then consider sum(X^n,n=1..). Since X < 1, this sum converges. But consider sum(1/(1+ip_1p_2...0_n),i=1..) Clearly sum(X^n,n=1..) > sum(1/(1+ip_1p_2...0_n),i=1..) as all terms are non-negative and every term in the rhs turns up in the lhs somewhere. But lim((1/(1+ip_1p_2..p_n))/(1/i),i=1..) = lim(i/(1+ip_1p_2..p_n)) = lim(i/ip_1p_2..p_n)=1/p_1..p_n is finite and nonzero. So since sum(1/i,i=1..) diverges, so does sum(1/(1+ip_1..p_n),i=1..), and thus so does sum(X^n,n=1..) which is a contradiction.
huh... have mathematicians studied the properties of similar objects where summation is done over all primes? i'm thinking (where p goes over all primes) Σ xᵖ Σ xᵖ/p! (maclaurin series of eˣ, but just the primes) Σ 1/xᵖ (does this converge? closed form?) Σ 1/p² (this should converge... right?)
Very neat proof. Although it did not need to be framed as a proof by contradiction. What you are showing is that every tail of the reciprocal of primes series has a sum X greater than 1, which you can show directly by, as in the video, showing that the geometric series X + X^2 + ... diverges. This is essentially the same proof but rewritten as a direct proof instead, which imo almost always makes things cleaner and clearer.
Does this only work because all the terms are positive? The series 1/2 -1/3 +1/4 -1/5 + 1/6... converges (not absolutely) but the subseries 1/2 +1/4 +1/6 .... is divergent. So it's not necessarily true that if the subseries diverges then the main series diverges as well.
Great video! There are a number of videos out there on the divergence of the harmonic series and the reciprocals of the prime numbers. Maybe there are some on generalising the harmonic series to the Riemann zeta function. The reciprocal of prime numbers can similarly be generalised to the prime zeta function. I have never seen a video on this! If this is something of interest, I can send a link to the Carl Froberg paper on the prime zeta function. It also has a wikipedia page, and I have been calculating its zeros for many years!
Isn't there some interpretation here that since 1/n^2 converges and 1/p diverges, there are "more primes" than squares? (Or at least, the primes occur more frequently than squares)
@@DrTrefor I think the concept he is alluding to is the concept of sparseness. Consider S to be some subset of N, and let X : N -> S be a bijective sequence of numbers in S. S is called a sparse set of N if the sequence of partial sums of 1/X converges. In other words, T = {n in N : m^2 = n, m in N} is a sparse set, since ζ(2) is finite, but P = {n in N : n is prime} is not a sparse set. Since geometric series converge, the set of powers of any given natural number is a sparse set. Etc.
I think one critical part of the proof has been left out: @8:00 the claim is that it's a subseries of the geometric series, but no proof has been given.
This was largely proved prior to the statement. That is I was arguing that the terms in the subseries would all look like the terms in the geometric series.
Is there a relationship between the reciprocal serie of diverging primes and the fact that the set of primes is infinite? why are we interested in the divergence of this serie?
Perhaps a simpler proof: Notice that p_i progresses like i*ln(i), and Sum(1/(i*log(i))) diverges for any log. We can check that that sum of 1/(i*log_2(i)) diverges by using the same trick as for the harmonic series. Group the first 2 terms, the next 4, the next 8, 16, 32, etc. In the harmonic series each group is >= 1/2 so the sum of the groups is infinite. In the 1/(i*log(i)) series, up to some multiplicative constant the first group is >= 1, the second is >= 1/2, the third is >= 1/3, etc. The sum of all the groups follows the harmonic series which diverges, so the 1/(i*log(i)) series diverges too.
That is what I initially thought too, but the chance is not 1/k, but rather 1/ln(k). Source: wikipedia (page 'prime number theorem') . Also see my comment (of a few minutes ago) to the video, where I do the integral.
@@koenth2359 Oh, i see, you are right. Probably got confused because I learned it from the computer science perspective and there you are using the length of the number rather than the number itself.
You could assume that since the reciprocal primes series is just a "portion" of the harmonic series (which is divergent), then the reciprocal primes series is divergent.
You can probably prove that for any convergent geometric series, the elements of that series, after a certain point, are always less than the corresponding elements of the reciprocal prime series.
Hmm, will need a few more view to digest it all, very nice. But I would have thought so, because we could approximate the result for large n by multiplying 1/n with the prime density function (which converges to 1/ln(n) for large n). If this sum is finite, the following integral should be finite too (I know, sloppy notation): ∞ ∫ 1/(x ln x) dx = ln(ln(∞)) - ln(ln(x0)) = ∞ x0
As with most (if not all?) proofs by contradiction, you didn't need to use contradiction. "Some series diverges and is a subseries of some geometric series with ratio X, so X >= 1. But X is an arbitrary tail of the reciprocal prime series so the reciprocal prime series diverges."
Don't you need contradiction to go from "arbitrary tail is >=1" to "the series converges"? (I define "series converges" as "for every M there is an initial part that exceeds M".) I guess one can use X>=1 to create disjoint initial parts of the series that are all >= 1/2, and when you take [M/2] + 1 of them, you can sum them to get >=M. But I personally prefer math without this hassle ;-) (even though I am a compatriot of LEJ Brouwer, and I am afraid no other Dutch mathematician compares to him).
@ronald3836 Show me any proof that proceeds "Assume B is not true and A is true... something something something... contradiction" and I will show you a proof "Suppose B is not true... something something something... therefore A is not true." It's not more hassle for me, except sometimes it forces me to understand things better. Because when I do a proof by contradiction, I have the sensation that I flailed around with symbols in Leprechaun and Unicorn-land until some contradiction fell on my lap, and I feel unenlightened by that exercise.
@@paradoxicallyexcellent5138 How do you prove the non-countability of the reals? Or do you simply not accept the statement that the reals are not countable?
@@ronald3836 Let f be any function from naturals to reals... something something diagonal... therefore f is not onto. At no point do I have to assume f is onto to prove it is not onto. This is merely a direct proof that "For all functions f from N to R, f is not onto." Assuming it is onto "for contradiction" is a farce.
I'm in junior level "theoretical methods" which is a math survey class at my university. We just covered series and how they can make/solve special functions like legendre, gamma, asymtopic series, etc Anyways, study group working on hw and we found a version of the p-series test where we would compare 1/f(n) whatever that is to 1/n^p where we're taking the limit at p-> 1+ So any value just to the right (greater) of p=1 and comparing. Does this have a name or anything? It's been very useful
7:50 say ... when you 'claimed' I went into a confused loop. Did you mean: "I will show that" ? I only ask because the flow of understanding is easier without tangential inputs. I mean, it was the first time you had mentioned 'any' claim in the first place, so it demanded a rerun of the video up until that point as a minimum.
The sum of twin primes is always divisible by 12 p>3 ,eg 41+43=84 = 12x7. Would think that would be used in proving the the convergence 1/3 + 1/5 + 1/5 + 1/7 + 1/11 + 1/13 + ….
I disagree with the stipulation that the sums of the reciprocal primes is greater than the sum of the reciprocals of powers of 2. I don't think you've proved this here.1:00
I’m effectively citing the fact there is always a prime between n and 2n, but just illustrating that for the first few terms as that isn’t the proof I’m aiming for in this video.
@@DrTrefor I think that would have been interesting in its own right. Given it's already an over 10 minute video, I don't see the need to skip it. For those of us in the know, the devil in infinite series is always in the detail and when you skip it, I don't think "that must have been boring". I think "why did he skip it?" "what's he hiding?". More importantly, I think you should be careful about the motivation for your content. Instead of curating maths, deciding what the public gets to see (and what it can't), consider instead that your content should be about you being a mathematician and showing the world what that is like and how you see things. Afterall, you're not a gatekeeper. You own none of this. That is the difference between Matt Parker and 3b1b and the rest of the horde of maths channels out there.
I think there is a skipped step at the end: what is actual proven is that any suffix of the original series is greater than 1. To be pedantic, you still need to prove that it's not a finite term greater than 1. (Intuition tells me that follows, but using intuition with infinities is a risky bet.) Also, why prove by contradiction? Wouldn't the argument work just as well if you start by proving that all the generated sub-series diverges, then show that implies that X is always grater than 1 for every possible suffix and conclude by showing that implies the original series diverges?
This does not require a formal proof. There are an infinite number of primes, therefore no matter how small the reciprocal becomes there are infinite more to add to the sum.
The sum of reciprocals of twin primes converges, defining the Braun constant. If Braun Constant are an irrational Number, then exists infinite twin primes.
Enjoyed. What about series made up of every k'th prime? For which values of k will they converge if any? You showed it diverges for k = 1, but what about k=2, the sum of the reciprocals of every second prime?
@@DrTrefor Sum of reciprocals of n'th primes never converges. Assume it does then we can generate a new series by adding the reciprocals of the next primes, i.e. those where p = 1 (mod n). This sum is less because all the denominators are larger. Similarly for the sum of all primes where p = a (mod n) for 0
Don't know why my other reply didn't pop up, but the key to all this sum-of-reciprocal stuff is the Bloom-Sisask theorem, which (in a sense) says that the sum of reciprocals of elements of A converges if the density of A is C/(logx)^(1+e) for some C>0 and e>0. As the density of primes is 1/logx, this tells us the sum of reciprocals of primes diverges. The subset of kth primes would have density porportional to the density of primes, which is x/logx, so that e mentioned in the theorem above will be 0, and the sum will still diverge. The theorem is actually about arithmetic progressions in A, but you can translate.
@@sk8erJG95 Had to look up density of a set of reals: en.wikipedia.org/wiki/Natural_density. I found a 2018 paper by Bloom and Sisak on arxiv but it's beyond my comprehension.
@@julianfogel5635 Look at Thomas Bloom's research page, he posts links to his papers and a small summary of the results of those paper. His summary: "We show that if A = {1,...,n} contains no 3-term arithmetic progression, then |A|
I wonder what the largest of the series of naturals (primes or otherwise) that converges is. I imagine you can get arbitrarily large, but what about restricting it to "relatively" simple patterns? That might be a really interesting question
@@DrTrefor Well, have you talked about the OEIS constants at A350581 or A350582? Would love to hear what you think about them. I find the difference between a bit weird. The question on my mind is what does this imply?
