(ln(n) * n)' = 1/(n^2) * (1 - ln(n))
which means when n = e = 2.718, the sequence is max. So it's the greater in a = 2^(1/2) and b = 3^(1/3).
a^6 = 8, b^6 = 9. So 3^(1/3).
which means when n = e = 2.718, the sequence is max. So it's the greater in a = 2^(1/2) and b = 3^(1/3).
a^6 = 8, b^6 = 9. So 3^(1/3).
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