Does the series ∑2∞ 1/ln n! converge? Does the series 1/3 + 1/(3 31/2) + 1/(3 31/2 31/3) + ... + 1/(3 31/2 31/3 ... 31/n) + ... converge?
Solution
(1) ln n! < n ln n, so ∑2∞ 1/ln n! > ∑2∞ 1/(n ln n), which diverges. For ∫ 1/(x ln x) = ln ln x.
(2) We have 1 + 1/2 + 1/3 + ... + 1/n is approx ln n, and 1/3ln nis approx 1/nln 3. But ln 3 > 1, so we expect this to converge. In fact 1 +1/2 + 1/3 + ... + 1/n > ∫1n+1 dx / x = ln(n + 1) > ln n. So 1/3 + 1/(3 31/2) + 1/(3 31/2 31/3) + ... + 1/(3 31/2 31/3 ... 31/n) + ... < 1/3ln 1 + 1/3ln 2 + 1/3ln 3 + ... = 1/1 + 1/2k + 1/3k + ... , where k = ln 3, which converges.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002