If you look at the first function, and pose
, well you can easily prove that
converges to 0, since it is only the harmonic sequence multiplied by a constant, and you notice that
={1, 0, 1, 0, 1 ...}
It can easily be shown that f(xn) diverges, since there exists an epsilon greater than zero(just pick smaller than 1) for which, no matter what rank N of the sequence
you pick, there exists i greater than N such that

is greater than epsilon(two successive terms in f(xn) always have a distance of exactly 1), therefore
is not Cauchy, which implies that it does not converge.