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In Ramanujan's proof of Bertrand's Postulate, Ramanujan states:

$\log([x]!) - 2\log([\frac{1}{2}x]!) \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$

where:

$\vartheta(x) = \sum_{p \le x} \log p$

$\psi(x) = \sum_{n\ge1} \vartheta(x^{\frac{1}{n}})$

$\log([x]!) = \sum_{n\ge1} \psi(\frac{1}{n}x)$

$\log([x!] - 2\log([\frac{1}{2}x]!) = \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x) - \psi(\frac{1}{4}x) + \ldots$

How does one go about proving that $\log([x]!) - 2\log([2x]!) \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$ is correct and something like $\log([2x!]) - \log([x!]) \le \psi(2x)$ is not correct?

Is $\log([3x!]) - \log([2x]!) \le \psi(3x) - \psi(2x) + \psi(\frac{3}{2}x)$?

Larry Freeman
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1 Answers1

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I found the answer to my question in M. Ram Murty, Problems in Analytic Number Theory.

If $a_0 \ge a_1 \ge a_2 \ge \ldots$ is a decreasing sequence of real numbers tending to $0$, then:

$a_0 - a_1 \le \sum_{n=0}^{\infty}(-1)^{n}a_n \le a_0 - a_1 + a_2$.

This result is straight forward to show using induction.

This can be applied to the Chebyshev second function since:

$\log([x]!) - 2\log([\frac{x}{2}]!) = \sum_{n \le x}(-1)^{n-1}\psi(\frac{x}{n})$

It is now straight forward to resolve $\log([2x]!) - \log([x]!)$ and $\log([3x!]) - \log([2x]!)$.

In both cases, the sequence may not be strictly decreasing so we cannot apply the upper bound.

In the case of $\log([2x]!) - \log([x]!)$, we have:

$\psi(2x) - \psi(x) + \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{2}{3}) - \ldots$ where $\psi(\frac{2}{3}x)$ may be greater than $\psi(\frac{1}{2}x)$.

In the case of $\log([3x]!) - \log([2x]!)$, we have:

$\psi(3x) - \psi(2x) + \psi(\frac{3}{2}x) - \psi(x) + \psi(x) - \psi(\frac{2}{3}x) + \psi(\frac{3}{4}x) - \ldots$ where $\psi(\frac{3}{4}x)$ may be greater than $\psi(\frac{2}{3}x)$.

Larry Freeman
  • 10,433