m>n
f(m) - f(n) = 1/n sum_{1<=k<=m}p{n/k} - 1/m sum_{1<= k <= m} p{m/k}, where p(x) = x - [x], the fractional part of x.
f(2n) > f(n) since p(x+y) <= p(x) + p(y)
This will prove 1
consider another case, n is a sufficiently large prime, m = p + 1
since in this p(m/k) = p(m/k) + p(1/k) when k < p-1
f(p+1) < f(p)
this will prove 2
f(m) - f(n) = 1/n sum_{1<=k<=m}p{n/k} - 1/m sum_{1<= k <= m} p{m/k}, where p(x) = x - [x], the fractional part of x.
f(2n) > f(n) since p(x+y) <= p(x) + p(y)
This will prove 1
consider another case, n is a sufficiently large prime, m = p + 1
since in this p(m/k) = p(m/k) + p(1/k) when k < p-1
f(p+1) < f(p)
this will prove 2
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