How about this:
n^(1/n) - 1 > 2^(1/n) - 1;
The definite integral of ( 2^(1/n) - 1 ) dn from 1 to infinity equals to
the definite integral of (2^x - 1)(-1/(x^2)) dx from 1 to 0 by making a substitution (x=1/n),
which, of course, equals to
the definite integral of ( (2^x )/(x^2) -...