The Reciprocal Prime Series (this proof should be taught in calculus!)

What is the sum of 1/p, where p is prime? There are infinitely many prime numbers, but as they become larger they contribute smaller and smaller amounts to the sum. So, does that sum converge or diverge? In this video I want to share a really cool proof that they diverge that will play on the harmonic series and the geometric series - to famous series from calculus - as well as prime factorization to prove that indeed this must diverge.

0:00 The Reciprocal Prime Series
0:30 The Harmonic and Geometries Series
2:37 The proof of divergence
10:27 Thanks to Maple Calculator

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[kasuha]

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.

[johnchessant]

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.

[changyauchen]

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.

[eofirdavid]

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.

[johnchessant]

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.

[cblpu]

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?

[alexismiller]

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

[hareal]

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) = ♾️

[robharwood]

@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.

[angelmendez-rivera]

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.

[geoperry]

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

[Jack_Callcott_AU]

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 .

[MasterHigure]

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<1 anywhere. It's a personal pet peeve of mine when people make assumptions for contradictions at the start of proofs, and then never actually use said assumptions until they reach the end, where they say "But we assumed ..." If that's the way you want to conclude, make the assumption at the end, don't make us keep tabs on it all the way through the proof.

That's different from just pointing out at the start that if the series converges, then there should be an N such that X<1. It's subtle, but the difference is there. Main distinction being, you can do the proof and the pointing out in either order, without any logic issues.

[zitengwang]

Finally getting someone is actually knowing math and teaching math properly on YouTube.

[victorhermestorrestomara]

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

[meurdesoifphilippe]

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.

[saggetarious]

At 2:34 the geometric series should start from i=0 for it to be 1/1-x, right? Very interesting video, thanks for all the great content! 😁

[sz]

Your explanation is so clear, thanks!

[ivailogeorgiev]

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.

[jerry]

Maple is an incredible tool, I hope someday every child will use it!