Mangoldt summatory function is $O(x)$

$\\psi(x)=O(x)$ , 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).

Mangoldt summatory function is O⁢(x)