0
$\begingroup$

So I stumbled upon this while trying to prove something else and cannot find a simple (read:elementary) proof for this (I'm old fashioned and still use $\log$ for natural log):

For large enough even $x$, $\log(8x+1)+\log(8x+3)+\cdots+\log(9x-1)-\log 1-\log 3-\cdots-\log(x-1)>\vartheta(9x)-\vartheta(8x)$.

For general $x$ as $\sum_{\substack{8x<k\leq9x\\k\text{ odd}}}\log k-\sum_{\substack{k\leq x\\k\text{ odd}}}\log k>\vartheta(9x)-\vartheta(8x)$.

I know it holds for at least $x\geq 11000$ by applying bounds to the Chebyshev function and using Stirling's approximation, but I would like to avoid using said Chebyshev bounds as they were proven with analytic results and I'm hoping for a more elementary approach.

Thanks in advance.

$\endgroup$
  • $\begingroup$ The trivial bound is $\displaystyle\vartheta(x+y)-\vartheta(x) = \sum_{p \in (x,x+y]} \log (p) \le \sum_{2n+1 \in (x,x+y]} \log (2n+1)$. Of course it is not true when $y$ is small that $\displaystyle\vartheta(x+y)-\vartheta(x) \le \sum_{2n+1 \in (x,x+y]} \log (2n+1)-\sum_{2n+1 \in [1,x]} \log (2n+1)$, but it will be true when $y/x$ is large enough. You can use $\sum_{n \le x} \log(n) \sim x \log x$. $\endgroup$ – reuns Aug 5 '17 at 22:23
  • $\begingroup$ Hey, reuns. Just a quick note that I edited the original question to a more specific form. Thank you. $\endgroup$ – Kyle Balliet Aug 22 '17 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.