6

$$\frac{1}{3}+\frac{1}{13}+\frac{1}{23}+\frac{1}{31}+\frac{1}{37}+\frac{1}{43}\cdots$$ Intuitively, I feel that this sum converges, but I really don't know why, (or if I am correct). Can I have a somewhat rigorous proof of whether or not this sum converges or diverges? Thank you lots for any help.

Apurv
  • 3,363
  • 2
  • 26
  • 47
  • its a collection of Dirichlet's – JMP Jul 16 '15 at 04:23
  • 5
    Any large number, and hence any large prime, is almost certain to contain the digit 3. Hence, apart from some initial segment, almost all primes appear in your series. – vadim123 Jul 16 '15 at 04:23
  • If you assume that asymptotically the number of primes with the digit three is equal to the number with three. This is because only the last digit of a prime is statistically "non-random". If you can prove that, you know the sum diverges. – Zach466920 Jul 16 '15 at 04:27
  • 2
    @AAron One in every $4$ primes ends in the digit $3$, so you should probably rethink your intuition about how small the sum is. – Erick Wong Jul 16 '15 at 05:08

2 Answers2

11

The sum of reciprocals of all positive integers without the digit $3$ converges. (See for instance Sum of reciprocals of numbers with certain terms omitted)

Hence the sum of reciprocals of all primes without the digit $3$ also converges.

But the sum of all prime reciprocals diverges.

Hence the sum of prime reciprocals with the digit $3$ must diverge.

Community
  • 1
paw88789
  • 40,402
0

By the Prime Number Theorem in arithmetic progressions (the extension of Dirichlet's Theorem), not only are there infinitely many primes which are 3 mod 10, but these primes are asymptotically 1/phi(10) = 1/4 of all primes. Thus, since the reciprocal sum of the primes diverges, so must the reciprocal sum of the primes which end in 3, albeit about 1/4 the 'speed'.

Of course this is just primes ending with 3; if you include all primes with 3s anywhere you'll quickly find that this is the vast majority of primes. For example, there are about 10^1000000 / (1000000 log 10) ≈ 4 * 10^999993 primes with up to a million digits, but only 9^1000000 ≈ 3 * 10^954242 numbers (prime or composite) with up to a million digits lacking 3s. That is, if you pick a random prime up to 10^1000000 the chance that it will have no 3s is less than 1 in 10^45751.

Charles
  • 32,122