Do the sum of all prime reciprocals with the digit $3$ converge or diverge? $$\frac{1}{3}+\frac{1}{13}+\frac{1}{23}+\frac{1}{31}+\frac{1}{37}+\frac{1}{43}\cdots$$
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.
 A: 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.
A: 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.
