3 ms·
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
by 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)) ...)).