upper bound on $\vartheta (n)$

Theorem.

Let $\vartheta \mathit{}\mathrm{(}n\mathrm{)}$ be the Chebyshev function

$$\vartheta (n)=\sum _{\begin{array}{c}p\le n\\ p\mathrm{prime}\end{array}}\log p.$$

Then $\vartheta \mathit{}\mathrm{(}n\mathrm{)}\mathrm{\le }n\mathit{}\log \mathit{}\mathrm{4}$ for all $n\mathrm{\ge }\mathrm{1}$.

Proof.

By induction.

The cases for $n=1$ and $n=2$ follow by inspection.

For even $n>2$, the case follows immediately from the case for $n-1$ since $n$ is not prime.

So let $n=2m+1$ with $m>0$ and consider ${(1+1)}^{2m+1}$ and its binomial expansion (http://planetmath.org/BinomialTheorem). Since $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m+1}}\right)$ and each term occurs exactly once, it follows that $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m}}\right)\le 4^m$. Each prime $p$ with divides $\left(\genfrac{}{}{0pt}{}{2m+1}{m}\right)$, implying that their product also divides $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+1}{m}}\right)$. Hence

$$\vartheta (2m+1)-\vartheta (m+1)\le \log \left(\genfrac{}{}{0pt}{}{2m+1}{m}\right)\le m\log 4.$$

By the induction hypothesis, $\vartheta (m+1)\le (m+1)\log 4$ and so $\vartheta (2m+1)\le (2m+1)\log 4$. ∎

Title	upper bound on $\vartheta (n)$
Canonical name	UpperBoundOnvarthetan
Date of creation	2013-03-22 16:09:47
Last modified on	2013-03-22 16:09:47
Owner	mps (409)
Last modified by	mps (409)
Numerical id	5
Author	mps (409)
Entry type	Theorem
Classification	msc 11A41