Log of factorial

$\sum_{n=1}^{x}\ln(n)=\ln(x!)$

We can approximate it's derivative by taking 2 secants equidistant (in the x-axis) from the point for which we have to approximate the derivative.

let $f(x)$ be derivative of $\ln(x!)$

$f(x)>\ln(x)$ ...(because secant from point of coordinates $(x-1,\ln((x-1)!))$)

$f(x)<\ln(x+1)$ ...(because secant from point of coordinates $(x+1,\ln((x+1)!))$)

As the difference between $\ln(x+1)$ and $\ln(x)$ decreases then we can more accurately approximate the value of $f(x)$

$\ln(x+1)-\ln(x)=\ln(\frac{x+1}{x})=\ln(1+\frac{1}{x})$

We know that as $x$ approaches infinity, $\frac{1}{x}$ approaches 0

=> As x approaches infinity $\ln(1+\frac{1}{x})$ approaches $\ln(1)=0$

=> $f(x)$ ~ $\frac{\ln(x)+\ln(x+1)}{2}$

$\frac{\ln(x)+\ln(x+1)}{2}=\frac{\ln(x(x+1))}{2}=2^{-1}\ln(x(x+1))=\ln(\sqrt{x(x+1)})$

So as $\sqrt{x(x+1)}$ ~ $x+\frac{1}{2}$ ...(1)

Proof of above statement (1) :

let say as $x$ approaches infinity $\sqrt{x(x+1)}$ approaches $x+k$ for any constant k

Therefore, $\sqrt{x(x+1)}=x+k$ must led to contradictions

let's simplify it

$x(x+1)=(x+k)^{2}$

$x^{2}+x=x^{2}+2xk+k^{2}$

$x=2kx+k^{2}$

if $k=\frac{1}{2}$ then $x=x+k^{2}$ => $k^{2}=0$

this led to contradiction

this mean that it only leads to contradiction only if $k=\frac{1}{2}$

this mean that $k$ must equal $\frac{1}{2}$

Hence $\sqrt{x(x+1)}$ ~ $x+\frac{1}{2}$

=> $f(x)$ ~ $\ln(x+\frac{1}{2})$

integrating both sides

$\ln(x!)$ ~ $∫\ln(x+\frac{1}{2})dx$

$∫\ln(x+\frac{1}{2})=(x+\frac{1}{2})(\ln(x+\frac{1}{2})-1)+c$

$\ln(x!)$ ~ $(x+\frac{1}{2})(\ln(x+\frac{1}{2})-1)+c$ ...(A)

$∏_{P≤N}P^{floor(\frac{N}{P})}=N!$ ...(here P is prime)

$floor(\frac{N}{P})$ ~ $\frac{N}{P}$

$\ln(N!)$ ~ $\sum_{P≤N}(\frac{N}{P})\ln(P)$ ...(B)

Combining (A) and (B)

$(N+\frac{1}{2})(\ln(N+\frac{1}{2})-1)+c$ ~ $N\sum_{P≤N}\frac{\ln(P)}{P}$

$\frac{(N+\frac{1}{2})(\ln(N+\frac{1}{2})-1)+c}{N}$ ~ $\sum_{P≤N}\frac{\ln(P)}{P}$

$(1+\frac{1}{2N})(\ln(N+\frac{1}{2})-1)+\frac{c}{N}$ ~ $\sum_{P≤N}\frac{\ln(P)}{P}$

$\ln(N+\frac{1}{2})-1$ ~ $\sum_{P≤N}\frac{\ln(P)}{P}$

$\ln(N)-1$ ~ $\sum_{P≤N}\frac{\ln(P)}{P}$

Note:

• Here, floor() is the floor function I used this notation because I was unable to find it's symbol.

• If you find any problem or difficulty in mathematics used here please comment.