# 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

Here are some additional inequalitys that might be helpful

$$\{ 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.

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math.stackexchange.com/users/47997/boby math.stackexchange.com/users/47996/boby math.stackexchange.com/users/47886/boby math.stackexchange.com/users/48220/boby You seem to have four different accounts. Kindly stick to one account and do not create new accounts. –  user17762 Nov 5 '12 at 2:55
I cant my accounts keep geting deleted, i cant log into my other account it says it doesnt exist or somthing it wont let me, i dont know where to ask 4 help? –  boby Nov 5 '12 at 2:57
If you plan on continuing to use MSE, I would recommend registering (look at the top right) so that you don't have to worry about cookies and caches to stay logged in. –  mixedmath Nov 5 '12 at 4:03
@boby Look here meta.stackoverflow.com/questions/81773/… on how to log out. –  user17762 Nov 5 '12 at 7:20
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