The n'th prime looks a bit like n log n. (This is basically the same fact as the "prime number theorem": the number of primes up to n is roughly n/log(n).)
We have:
sum (up to n) of 1/x ~= log n [harmonic series]
sum (up to n) of 1/(x log x) ~= log log n [reciprocals of primes, roughly]
sum (up to n) of 1/(x log x log log x) ~= log log log n
and so on. All these series diverge, but slower and slower and slower. And clearly these series "only just" diverge. In fact, iIf you take, e.g.,
sum of 1/(x log x log log x (log log log x)^1.1)
then it converges. If you have just the right kind of warped mind, the question that will immediately occur to you on contemplating this is: what if you define L(x) := x log x log log x log log log x ... continuing the product up to, but excluding, the first factor that's < 1, and ask about the sum of 1/L(x)?
Answer: The series diverges really slowly. The points at which L(x) starts to use one extra logarithm are e, e^e, e^e^e, e^e^e^e, etc. The sums between each adjacent pair of these are comparable in size. So the sum up to n is roughly the number of times you need to take logs, starting with n, before you get down to 1.
[EDITED to add: The first bit of the above is closely related to cperciva's comment that's a sibling of this one.]
If you've never tried Project Euler, it's a great way to learn new mathematical concepts, sequences, dynamic programming and other algorithms.
https://projecteuler.net/problem=368 was one of my favorites, because it involved turning an algorithm described in a paper into code.
I don't consider this a spoiler for problem #500, but if you're counting the number of divisors, you'll need this function: http://oeis.org/wiki/Number_of_divisors_function#Formulae_fo...
As an example, 120 = 2^3 * 3^1 * 5^1
number of divisors of 120 => (3+1)(1+1)(1+1)
Even if you don't solve it, exploring the problem is a fun afternoon for a person with a computer.