Mangoldt summatory function is $O(x)$

Theorem 1

$\psi(x)=O(x)$ , in other , $\frac{\psi(x)}{x}$ is bounded.

Proof.

$\psi(x)=\sum_{1}^{x}\Lambda(n)=\sum_{\begin{subarray}{c}p\text{ prime}\\ p\leq x\end{subarray}}\lfloor\log_{p}x\rfloor\ln p=\sum_{\begin{subarray}{c}p% \text{ prime}\\ p\leq x\end{subarray}}\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p=\sum_{% \begin{subarray}{c}p\text{ prime}\\ p\leq\sqrt{x}\end{subarray}}\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p+% \sum_{\begin{subarray}{c}p\text{ prime}\\ \sqrt{x}<p\leq x\end{subarray}}\ln p$

since $1\leq\frac{\ln x}{\ln p}<2$ if $p>\sqrt{x}$ . Continuing, we have

$\sum_{\begin{subarray}{c}p\text{ prime}\\ p\leq\sqrt{x}\end{subarray}}\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p+% \sum_{\begin{subarray}{c}p\text{ prime}\\ \sqrt{x}<p\leq x\end{subarray}}\ln p\leq\sqrt{x}\ln x+\pi(x)\ln x\leq\sqrt{x}% \ln x+8x\ln 2=O(x)$

Note that $\pi(x)\ln x\leq 8x\ln 2$ by Chebyshev’s bounds on $\pi(x)$ (http://planetmath.org/BoundsOnPin).

Title	Mangoldt summatory function is $O(x)$
Canonical name	MangoldtSummatoryFunctionIsOx
Date of creation	2013-03-22 17:42:59
Last modified on	2013-03-22 17:42:59
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	5
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11A41

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Theorem 1

Mangoldt summatory function is $O(x)$