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Can someone find a function f(n) satisfieing these bounds? Can you also prove that it does? $ \sum\limits_{k=1}^n \Lambda(k) [1-\text{Frac}(\frac{n}{k})][1-\frac{k}{n}\text{Frac}(\frac{n}{k})]=\frac{1}{2}\sum\limits_{k=1}^n \Lambda(k){}\text{}+O(f(n)),\text{ Such that:}\lim_{n\to\infty}f(n)/n=0$

Where $\displaystyle \text{Frac}(\frac{n}{k})$ is the fractional part of $\displaystyle \frac{n}{k}$, and where $\Lambda(k)$ is the Von-Mangoldt function. I know that a function does exist, I just cant prove that it does. I would Greatly appreiciate any help though, and if someone could even give me an elementary proof I would be willing to do somthing for them in return.

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I doubt there is a reasonable elementary proof. Note that the left-hand sum equals $ \sum_{F=1}^n \sum_{n/(F+1) If we let $\psi(x) = \sum_{1\le n\le x} \Lambda(n)$, then the inner sum can be written as a Riemann-Stieltjes integral $ \sum_{n/(F+1) which after integration by parts equals \begin{align*} \sum_{n/(F+1)$ \frac1n \sum_{F=1}^n F \bigg( \psi(\tfrac nF) - (F+1) \int_{n/(F+1)}^{n/F} \psi(x) \,dx \bigg). $ So far this is all elementary; and if you define $E(x) = \psi(x) - x$, then this becomes \begin{multline*} \frac1n \sum_{F=1}^n F \bigg( \tfrac nF - (F+1) \int_{n/(F+1)}^{n/F} x \,dx \bigg) + \frac1n \sum_{F=1}^n F \bigg( E(\tfrac nF) - (F+1) \int_{n/(F+1)}^{n/F} E(x) \,dx \bigg) \\ = \frac12 + \frac1n \sum_{F=1}^n F \bigg( E(\tfrac nF) - (F+1) \int_{n/(F+1)}^{n/F} E(x) \,dx \bigg) \end{multline*} after some easy evaluation.

However, I don't know how to proceed further without plugging in the Prime Number Theorem, say in the form $|E(x)| \le C(A) x/(\log x)^A$ for any $A>0$ and some constant $C(A)$ depending on $A$.

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    $F$ is simply an index of summation (see the $\sum_{F=1}^n$?). In the first step, it equals the integer part of $n/k$; thereafter, it's simply an integer parameter.2012-11-03