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[parent] Mangoldt summatory function is $O(x)$ (Theorem)
Theorem 1   $\psi(x)=O(x)$ in other words, $\frac{\psi(x)}{x}$ is bounded.
Proof. $$\psi(x)=\sum_1^x \Lambda(n)= \sum_{\substack{p\text{ prime}\\p\leq x}}\lfloor \log_p x\rfloor \ln p= \sum_{\substack{p\text{ prime}\\p\leq x}}\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p= \sum_{\substack{p\text{ prime}\\p\leq \sqrt{x}}}\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p + \sum_{\substack{p\text{ prime}\\\sqrt{x}<p\leq x}}\ln p$$ since $1\leq \frac{\ln x}{\ln p}<2$ if $p>\sqrt{x}$ Continuing, we have $$\sum_{\substack{p\text{ prime}\\p\leq \sqrt{x}}}\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p + \sum_{\substack{p\text{ prime}\\\sqrt{x}<p\leq x}}\ln p \leq \sqrt{x}\ln x+\pi(x)\ln x\leq\sqrt{x}\ln x+8x\ln 2=O(x) $$ Note that $\pi(x)\ln x\leq 8x\ln 2$ by Chebyshev's <</A>49#>bounds on $\pi(x)$ http://planetmath.org/encyclopedia/BoundsOnPin.html.




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Cross-references: proof, bounded

This is version 2 of Mangoldt summatory function is $O(x)$, born on 2007-12-28, modified 2008-05-24.
Object id is 10161, canonical name is MangoldtSummatoryFunctionIsOx.
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AMS MSC11A41 (Number theory :: Elementary number theory :: Primes)

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