PlanetMath: Mangoldt summatory function is $O(x)$

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Mangoldt summatory function is $O(x)$

(Theorem)

Theorem 1 $\psi(x)=O(x)$ , in other words, $\frac{\psi(x)}{x}$ is bounded.

Proof.

$\displaystyle \psi(x)=\sum_1^x \Lambda(n)= \sum_{\substack{p\text{ prime}\\ p\l... ...p}\right\rfloor\ln p + \sum_{\substack{p\text{ prime}\\ \sqrt{x}<p\leq x}}\ln p$

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

$\displaystyle \sum_{\substack{p\text{ prime}\\ p\leq \sqrt{x}}}\left\lfloor\fra... ...x}<p\leq x}}\ln p \leq \sqrt{x}\ln x+\pi(x)\ln x\leq\sqrt{x}\ln x+8x\ln 2=O(x)$

Note that $\pi(x)\ln x\leq 8x\ln 2$ by Chebyshev's bounds on $\pi(x)$ .

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Cross-references: 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.
Accessed 269 times total.

Classification:

AMS MSC:

11A41 (Number theory :: Elementary number theory :: Primes)

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