Proving a simple inequality

Can someone show that the inequality bellow holds? $$f(n) \leq f(n+1) \$$ Where $$\frac{\sum\limits_{k=1}^n \Lambda(k) {k}/{n}\lceil{n}/{k}\rceil{}\{ n/k \}}{\sum\limits_{k=1}^n \Lambda(k)}=f(n)$$

It may seem complicated, but I don't think it is

$$\{ n/k \} \leq (k-1)/n \$$
Where { n/k } is the fractional part of n/k , $\lceil{n}/{k}\rceil{}$ is the ceiling function applied to n/k, and where $\Lambda(k)$ is the Von-Mangoldt function.
I find $$f(3) = \frac{\tfrac{2}{3}\log 2}{\log 6} = 0.2579... > f(4) = \frac{\tfrac{1}{2}\log 3}{\log 12} = 0.2210...$$ Are you sure the inequality is correctly written? –  Esteban Crespi Nov 5 '12 at 17:30