Bounding a Von-Mangoldt Summatory Function

Question

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.

Answer 1

I doubt there is a reasonable elementary proof. Note that the left-hand sum equals $$ \sum_{F=1}^n \sum_{n/(F+1)<k\le n/F} \Lambda(k)[1-(\tfrac nk-F)][1-\tfrac kn(\tfrac nk-F)] = \frac1n \sum_{F=1}^n F \sum_{n/(F+1)<k\le n/F} \Lambda(k)[(F+1)k-n]. $$ 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 $$ \sum_{n/(F+1)<k\le n/F} \Lambda(k)[(F+1)k-n] = \int_{n/(F+1)}^{n/F} [(F+1)x-n] \, d\psi(x), $$ which after integration by parts equals \begin{align*} \sum_{n/(F+1)<k\le n/F} \Lambda(k)[(F+1)k-n] &= [(F+1)x-n]\psi(x) \bigg|_{n/(F+1)}^{n/F} - \int_{n/(F+1)}^{n/F} \psi(x) \tfrac d{dx}[(F+1)x-n] \,dx \\ &= \psi(\tfrac nF) - (F+1) \int_{n/(F+1)}^{n/F} \psi(x) \,dx. \end{align*} (And this formula, once suspected, can be proved in an elementary fashion without needing to use Riemann-Stieltjes integrals.) In summary, your left-hand sum equals $$ \frac1n \sum_{F=1}^n F \bigg( \psi(\tfrac nF) - (F+1) \int_{n/(F+1)}^{n/F} \psi(x) \,dx \bigg). $$ 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 $A$.