# Ramanujan Proof of Bertrand Postulate, first step

Here is a link to his proof: http://www.zyymat.com/ramanujans-proof-of-bertrands-postulate.html.

I understand the first two "steps", that $$\vartheta(x)=\log2+\log3+\ldots+\log p$$ where $p$ is the greatest prime less than or equal to $x$. Step two is that $$\psi(x)=\vartheta(x)+\vartheta(x^{1/2})+\vartheta(x^{1/3})+\ldots$$ And then I read that \begin{align}\log(\lfloor x\rfloor !)&=\vartheta(x)+\vartheta(x^{1/2})+\vartheta(x^{1/3})+\ldots\\ &+\vartheta(\frac 12 x)+\vartheta(\frac 12 x^{1/2})+\vartheta(\frac 12x^{1/3})+\ldots\\ &+\vartheta(\frac 13 x)+\vartheta(\frac 13 x^{1/2})+\vartheta(\frac 13x^{1/3})+\ldots\\ &+\ldots \end{align} And I was like "Woah!!! Where did that come from?!"
So where does it come from?

Consider the Von Mangoldt function $\Lambda$ which satisfies the following identities $$\psi(x):=\sum_{n\leq x} \Lambda(n)=\ln(\operatorname{lcm}(1,2,3,\dots\lfloor x\rfloor))\quad\text{and}\quad \ln(n)=\sum_{d\mid n}\Lambda(d)$$ where $\psi(x)$ is the so-called second Chebyshev function. Then \begin{align}\exp\left(\sum_{k\geq 1}\vartheta(x^{1/k})\right)&=\prod_{k\geq 1}\exp\left(\vartheta(x^{1/k})\right)=\prod_{k\geq 1}\prod_{p\leq x^{1/k}}p\\ &=\prod_{k\geq 1}\prod_{p^k\leq x}p=\operatorname{lcm}(1,2,3,\dots\lfloor x\rfloor)=\exp(\psi(x)) \end{align} where $\vartheta(x):=\sum_{p\leq x}\ln(p)$ is the so-called first Chebyshev function. Moreover \begin{align}\log(\lfloor x\rfloor !)&=\sum_{n\leq x} \ln(n)=\sum_{n\leq x} \sum_{d\mid n}\Lambda(d)\\ &=\sum_{dk\leq x} \Lambda(d)=\sum_{k\leq x}\sum_{d\leq x/k} \Lambda(d)\\ &=\sum_{1\leq k\leq x}\psi(x/k)=\sum_{k\geq 1}\psi(x/k). \end{align}
• Why does $$\psi(x) = \sum_{n\leq x} \Lambda(n)$$? – D.R. Nov 3 '17 at 13:50
• @D.R. This is the definition of the second Chebyshev function $\psi(x)$. See en.wikipedia.org/wiki/Chebyshev_function – Robert Z Nov 3 '17 at 13:55