Summation of $\log\left(\frac i2\right)$

Question

Could someone just briefly explain why this summation is true?

$$\sum_{i = 1}^n\log\left(\frac i2\right) = \frac n2(\log n - 1)$$

I'm having a hard time wrapping my head around this. Any help is appreciated, thanks.

Answer 1

$$\sum_{i=1}^n \log(i/2) = \log \prod_{i=1}^n (i/2) = \log \frac{n!}{2^n} = \log(n!) - n \log 2.$$

Stirling's approximation (applied somewhat crudely) gives $\log(n!) \approx n \log n - n$, so $$\sum_{i=1}^n \log(i/2) \approx n \log n - (1 + \log 2) n = n(\log n - (1+\log 2)).$$

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If the original quantity is instead slightly different (the expression in your original screenshot is ambiguous), then $$\frac{1}{2}\sum_{i=1}^n \log i = \frac{1}{2} \log (n!) \approx \frac{1}{2} (n \log n - n) = \frac{n}{2} (\log n - 1).$$

Again, both of these are approximations (with an error term on the order of $\log n$), not equalities.