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I know that better and better bounds of the Chebyshev Theta and Psi functions are implied by knowing that the first (insert large number here) zeta zeroes lie on the Critical Line. These bounds, specifically for $\theta(n)$, seem to approach the bound of $\theta(n) \leq n$ for all $n \in \mathbb{N}$.

  1. Is this bound known to be true or false? I haven't found anything specifically addressing the topic, so I should probably start here.
  2. Is this bound known to imply/be implied by the Riemann Hypothesis? Since they seem so interwoven it only seems logical that one would imply the other.
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  • $\begingroup$ $\frac{\zeta'(s)}{s\zeta(s)} + \frac{1}{s-1} = \int_1^\infty (x-\psi(x)) x^{-s-1} dx$. the basic Mellin transform properties tell you that $x-\psi(x) \sim \sum_{Re(\rho) = \sigma_0} \frac{x^{\rho}}{\rho}$ where $\sigma_0=1/2$ is the Riemann hypothesis. the fact that all these poles are non reals implies that $x-\psi(x)$ changes of sign infinitely many times. $\endgroup$
    – reuns
    Commented Feb 15, 2016 at 8:06

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Although Dusart has shown that $\vartheta(x)<x$ for all $x<8\cdot 10^{11}$, it is not true in general that $\vartheta(x)<x$. Indeed, Littlewood’s oscillation theorem implies that there is a positive constant $c$ and infinitely many $x$ for which $$\vartheta(x)>x+c\sqrt{x}\log\log\log(x).$$

For more information on this, see Dietrich Burde's answer to this question on math.stackexchange and the cited references.

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The proposed bound is known to be false. Littlewood proved that $\theta(n)-n$ is infinitely often as large as a constant times $\sqrt n\log\log\log n$ (and as small as a negative constant times the same quantity). A proof appears for example in Montgomery and Vaughan's book.

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  • $\begingroup$ Thanks so much! Do you happen to know of a place where I can find a proof of this without having to buy a separate book? $\endgroup$
    – Carl Schildkraut
    Commented Oct 28, 2015 at 3:26
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    $\begingroup$ (It should be \sqrt{n}\log\log\log{n}, of course.) $\endgroup$
    – alpoge
    Commented Oct 28, 2015 at 3:29
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    $\begingroup$ Since it is false it does indeed imply the Riemann Hypothesis. $\endgroup$
    – Jim Conant
    Commented Oct 28, 2015 at 11:02

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