TITLE: How to calc $\log1+\log2+\log3+\log4+...+\log N$?
QUESTION [5 upvotes]: How to calculate $\log1+\log2+\log3+\log4+...+\log N= log(N!)$? Someone told me that it's equal to $N\log N,$ but I have no idea why.

REPLY [4 votes]: $$x-1\le\lfloor x\rfloor\le x$$ so that
$$\log(x-1)\le\log\lfloor x\rfloor\le\log x$$
and
$$\int_0^n\log x\,dx<\sum_{k=1}^n\log k<\int_{1}^{n+1}\log x\,dx.$$
Hence
$$n\log n-n<\log n!<(n+1)\log(n+1)-n.$$
For instance, with $n=100$,
$$360.5170185988\cdots<363.7393755555\cdots<366.1271722009\cdots$$

A better approximation is given by the famous Stirling's formula,
$$n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n,$$
or 
$$\log n!=n\log n-n+\frac12\log n+\frac12\log2\pi.$$
For $n=100$,
$$363.7385422250\cdots$$