Analytic Number Theory: Problem in Bertrand’s postulate I am trying to learn Bertrand’s postulate. I can not understand two steps


*

*Why $\displaystyle\sum_{n \leq x}\log n=\sum_{e \leq x} \psi\left(\frac{x}{e}\right)$, 
where $\psi(x)=\displaystyle\sum_{p^\alpha \leq x, \alpha \geq 1}\log p$?

*$\displaystyle\sum_{n \leq x}\log n- 2\displaystyle\sum_{n \leq x/2}\log n\leq \psi(x)-\psi(x/2)+\psi(x/3)$. 


Will you kindly help me. 
 A: This question has an answer to the first part of your question. Chebyshev's original proof is online and you don't have to rely on my translation of his argument. In case the link breaks, it is Vol. I of Chebyshev's Oeuvres, p. 49, Memoire Sur Les Nombres Premiers at p. 53. 
For the second part, begin with
$\hspace{55mm}\log [x]! = \sum \log x =$
$$
\begin{Bmatrix} 
 ~\theta(x)~~~ +~~~\theta(x)^{1/2}~~~+\theta(x)^{1/3}~~+ ~~...\\ 
 +~\theta(x/2)+\theta(x/2)^{1/2}+\theta(x/2)^{1/3}+...\\
 +~\theta(x/3)+\theta(x/3)^{1/2}+\theta(x/3)^{1/3}+...\\ 
\end{Bmatrix}
$$
$=\psi(x)+\psi(x/2)+\psi(x/3)+...=\sum_{e\leq x}\psi(x/e),$
in which $\theta(x) = \sum_{p\leq x} \log p,$ and because it's easier to look at I put $\theta(x/3)^{1/2}$ for $\theta((x/3)^{1/2}).$ 
If this seems laborious it makes your second inequality obvious. 
$\sum_{n\leq x} \log n - 2 \sum_{n\leq x/2} \log n = $
$$
\begin{Bmatrix} 
 ~\theta(x)~~~ +~~~\theta(x)^{1/2}~~~+\theta(x)^{1/3}~~+ ~~...\\ 
 +~\theta(x/2)+\theta(x/2)^{1/2}+\theta(x/2)^{1/3}+...\\
 +~\theta(x/3)+\theta(x/3)^{1/2}+\theta(x/3)^{1/3}+...\\ 
\end{Bmatrix}
$$
minus
$$
\begin{Bmatrix} 
 ~2 \theta(x/2)~~~ +~~~2\theta(x/2)^{1/2}~~~+2\theta(x/2)^{1/3}~~+ ~~...\\ 
 +~2\theta(x/4)+2\theta(x/4)^{1/2}+2\theta(x/4)^{1/3}+...\\
 +~2\theta(x/6)+2\theta(x/6)^{1/2}+2\theta(x/6)^{1/3}+...\\ 
\end{Bmatrix}
$$
which is equal to
$$A=
\begin{Bmatrix} 
 ~\theta(x)~~~ +~~~\theta(x)^{1/2}~~~+\theta(x)^{1/3}~~+ ~~...\\ 
- ~\theta(x/2)-\theta(x/2)^{1/2}-\theta(x/2)^{1/3}-...\\
+ ~\theta(x/3)+\theta(x/3)^{1/2}+\theta(x/3)^{1/3}+...\\ 
\end{Bmatrix}
$$
plus
$$B=
\begin{Bmatrix} 
- ~\theta(x/4)~~~ -~~~\theta(x/4)^{1/2}~~~-\theta(x/4)^{1/3}~~- ~~...\\ 
+ ~\theta(x/5)+\theta(x/5)^{1/2}+\theta(x/5)^{1/3}+...\\
- ~\theta(x/6)-\theta(x/6)^{1/2}-\theta(x/6)^{1/3}-...\\
+ ~\theta(x/7)+\theta(x/7)^{1/7}+\theta(x/7)^{1/3}+...\\
\end{Bmatrix}
$$
Note that $A \leq \psi(x)-\psi(x/2)+\psi(x/3) .$  Then note that in B the subtracted row is always greater than the the subsequent added row, 
so that the sum $A+ B \leq  \psi(x)-\psi(x/2)+\psi(x/3).$
