# Bounding a Von-Mangoldt Summatory Function

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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## 1 Answer

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 $A$.

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If I could reduce the prime number theorem to this problem, do you think it would be worth pursuing? Also are Riemann-Stieltjes integrals considered elementary? – boby Oct 26 '12 at 9:55
Personally, I think this problem is slightly harder than the prime number theorem (where the summand is simply $\Lambda(k)$). So reducing PNT to this problem isn't likely to be fruitful in my opinion. Riemann-Stieltjes integrals are more or less as elementary as regular Riemann integrals - they're just less common. The good news is, the formula I derived using R-S integrals can be verified using nothing more than first-year calculus. (Also, "elementary" isn't a firm label, so deciding whether a method is or isn't elementary isn't that mathematically significant.) – Greg Martin Oct 26 '12 at 21:29
What is meant by the letter F? does that denote the fractional part of n/k? or what? – user47704 Oct 31 '12 at 18:49
$F$ 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. – Greg Martin Nov 3 '12 at 5:38