How do you prove that #sum_(n=1)^oo (n^(1/n)-1)# diverges?
1 Answer
Feb 12, 2017
Use a comparison test against the harmonic series.
Explanation:
Use a comparison against the harmonic series:
#sum_(n=1)^oo 1/n#
which is known to diverge.
Note that:
#e^x = 1+x/(1!)+x^2/(2!)+x^3/(3!)+...#
#n^(1/n) = e^(1/n ln n) = 1 + (1/n ln n)/(1!) + (1/n ln n)^2/(2!) + ...#
When
#n^(1/n) - 1 = (1/n ln n)/(1!) + (1/n ln n)^2/(2!) + ... > 1/n#
Hence:
#sum_(n=3)^oo (n^(1/n)-1) > sum_(n=3)^oo 1/n#
diverges.