4 ms·

One of the things that's freaky to me is that the harmonic series up to n terms is proportional to log(n)... but the sum of the reciprocals of the primes is ro

by ipsin 12y ago

One of the things that's freaky to me is that the harmonic series up to n terms is proportional to log(n)... but the sum of the reciprocals of the primes is roughly proportional to log(log(n)). That's a crazy slow divergence.

cperciva 12y ago
For even more crazy, consider the following series: H(n) = 1/1 + 1/2 + ... + 1/n H_2(n) = 1 / (H(1)) + 1 / (2 H(2)) + 1 / (3 H(3)) + ... + 1 / (n H(n)) H_3(n) = 1 / (H(1) H_2(1)) + 1 / (2 H(2) H_2(2)) + 1 / (3 H(3) H_2(3)) + ... + 1 / (n H(n) H_2(n)) ... H_k(n) = 1 / (H(1) H_2(1) ... H_{k-1}(1)) + 1 / (2 H(2) H_2(2) ... H_{k-1}(2)) + ... + 1 / (n H(n) H_2(n) ... H_{k-1}(n)) The series H_k(n) diverges roughly proportional to log(log(... k logs ... (log(n)) ...)).
gjm11 12y ago
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.]

by ipsin 12y ago

cperciva 12y ago

For even more crazy, consider the following series: H(n) = 1/1 + 1/2 + ... + 1/n H_2(n) = 1 / (H(1)) + 1 / (2 H(2)) + 1 / (3 H(3)) + ... + 1 / (n H(n)) H_3(n) = 1 / (H(1) H_2(1)) + 1 / (2 H(2) H_2(2)) + 1 / (3 H(3) H_2(3)) + ... + 1 / (n H(n) H_2(n)) ... H_k(n) = 1 / (H(1) H_2(1) ... H_{k-1}(1)) + 1 / (2 H(2) H_2(2) ... H_{k-1}(2)) + ... + 1 / (n H(n) H_2(n) ... H_{k-1}(n)) The series H_k(n) diverges roughly proportional to log(log(... k logs ... (log(n)) ...)).

gjm11 12y ago

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.]