Reasoning about the Chebyshev functions: How does one check an upper bound based on the second Chebyshev function? 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 $\log([3x!]) - \log([2x]!) \le \psi(3x) - \psi(2x) + \psi(\frac{3}{2}x)$?

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 $\psi(\frac{2}{3}x)$.  
