Log of factorial


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

f(x)>ln(x)f(x)>\ln(x) ...(because secant from point of coordinates (x1,ln((x1)!))(x-1,\ln((x-1)!)))

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

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


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

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

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


So as x(x+1)\sqrt{x(x+1)} ~ x+12x+\frac{1}{2} ...(1)

Proof of above statement (1) :

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

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

let's simplify it




if k=12k=\frac{1}{2} then x=x+k2x=x+k^{2} => k2=0k^{2}=0

this led to contradiction

this mean that it only leads to contradiction only if k=12k=\frac{1}{2}

this mean that kk must equal 12\frac{1}{2}

Hence x(x+1)\sqrt{x(x+1)} ~ x+12x+\frac{1}{2}

=> f(x)f(x) ~ ln(x+12)\ln(x+\frac{1}{2})

integrating both sides

ln(x!)\ln(x!) ~ ln(x+12)dx∫\ln(x+\frac{1}{2})dx


ln(x!)\ln(x!) ~ (x+12)(ln(x+12)1)+c(x+\frac{1}{2})(\ln(x+\frac{1}{2})-1)+c ...(A)

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

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

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

Combining (A) and (B)

(N+12)(ln(N+12)1)+c(N+\frac{1}{2})(\ln(N+\frac{1}{2})-1)+c ~ NPNln(P)PN\sum_{P≤N}\frac{\ln(P)}{P}

(N+12)(ln(N+12)1)+cN\frac{(N+\frac{1}{2})(\ln(N+\frac{1}{2})-1)+c}{N} ~ PNln(P)P\sum_{P≤N}\frac{\ln(P)}{P}

(1+12N)(ln(N+12)1)+cN(1+\frac{1}{2N})(\ln(N+\frac{1}{2})-1)+\frac{c}{N} ~ PNln(P)P\sum_{P≤N}\frac{\ln(P)}{P}

ln(N+12)1\ln(N+\frac{1}{2})-1 ~ PNln(P)P\sum_{P≤N}\frac{\ln(P)}{P}

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


  • 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.

Note by Zakir Husain
1 year ago

No vote yet
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