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Chebyshev first proved that there exist constants $a$, $b$ such that $$ a \frac{n}{\log n} < \pi(n) < b \frac{n}{ \log n}.$$ The proof is well understood, and relies on elementary techniques.

However, Sylvester extended this proof, and according to this website , the explicit bound holds with $b = 1.04423$ for large enough $n$. I would be grateful if someone could provide a proof of this claim, or a sketch of the methods used by Sylvester

Navvye
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  • Did you search for Sylvester's paper? Later Dusart showed a much better estimate with much smaller $b>1$ (at the cost of larger $n$ for which this holds) - see your link or this post. – Dietrich Burde Apr 15 '24 at 08:41
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    I find it more interesting to have a reasonably good value for the bound $b>1$, where the inequality holds for all $n>n_0$, with a relatively small $n_0$. For $b=1.10555042752$ being Chebyshev’s constant, one obtains $n_0=96098$, see my paper here. – Dietrich Burde Apr 15 '24 at 08:46
  • @DietrichBurde I read your paper before posting, it was very interesting and I agree that having a reasonably good value for $b$ is better when it holds for all small enough $n_0$. However, in this case, I need $b$ to be less than $\ln 3$, as I'm using it as a lemma which is a part of a more complicated theorem that I'm trying to prove. Also, I did search for Sylvester's paper: "J.J. Sylvester, On arithmetical series, Messenger of Math", but could not find it. – Navvye Apr 15 '24 at 09:34
  • Also, I would appreciate if the bound is exactly of the form $\pi(n) < b \frac{n}{\log n}$ with the numerator and denominator untouched. – Navvye Apr 15 '24 at 09:38
  • From Dusart's result $\pi(x)<\frac{x}{\log(x)}\left( 1+\frac{1}{\log(x)}+\cdots \right)$ you can derive the inequalities you need (or would like to show). No need to look for older papers here, if you are not interested in the history of these results. – Dietrich Burde Apr 15 '24 at 11:35
  • I see, thank you! But I'm still interested in how Sylvester extended Chebyshev's results to obtain the bound, and I'm wondering if we can obtain a better bound using the same techniques and recent advancements in analytic NT – Navvye Apr 15 '24 at 12:43
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    I suppose you can ask a university library for Sylvester's paper. On the other hand, Dusart's arguments are also elementary and will give you the same. – Dietrich Burde Apr 15 '24 at 12:55
  • x * 1 / log x underestimates pi(x). You'd have to replace the 1 with 1 + 1 / ln x + 1 / ln^2 x.... Any constant 1 + 1/d instead of 1 overestimates pi(x) for x >= approximately exp(d). So you can easily find different constants giving upper bounds for different ranges. Or you can use something like 1.1 * x / ln x + 65 which is quite close and valid for all x. Just start up your computer and let it search. You can find different results, but nothing better than this one. – gnasher729 Nov 07 '24 at 17:24

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