How do you prove that #sum_(n=1)^oo (n^(1/n)-1)# diverges?

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Answer 1 · 2017-02-12T20:35:11

Use a comparison test against the harmonic series.

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 >= 3# we have #ln n > 1#, so:

#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.