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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.

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4 Answers

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$$x-1\le\lfloor x\rfloor\le x$$ 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 $n=100$,

$$360.5170185988\cdots<363.7393755555\cdots<366.1271722009\cdots$$

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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 $n=100$,

$$363.7385422250\cdots$$

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A small $caveat$: "someone" is wrong: try, for example, with $N=2$; then

$$2\log 2=\log 2^2,$$

while

$$\log1+\log 2=\log 2!=\log 2.$$

For more details on the relationship between the 2 logarithms, I refer to the comments under the OP.

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This holds true only for large values of N and Asymptotic notation. i.e. $log(N!)=O(NlogN).$

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The integral

$$\int_1^{N}\log(x) \; dx= N\log N - N +1$$

is a good approximation.

For $N=100$, the sum equals $361.3197940$ while the above expression gives $361.5170185.$

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