TITLE: How to show $\sum\limits_{i=1}^n\log\left(\frac{n}i\right)= \Theta(n)$?
QUESTION [6 upvotes]: Is the sum from i=1 to n for log(n/i) = Θ(n)?
Im studying for a test and appreciate your help.
This is what I did: and got something else
$$\sum_{i=1}^n \log(n/i)=\sum_{i=1}^n[\log n-\log i]=\left(\sum_{i=1}^n\log n\right)-\left(\sum_{i=1}^n \log i\right)=n\log n-\log n!=\log\left(\frac{n^n}{n!}\right) $$

REPLY [9 votes]: You did it all by yourself, really: 

Stirling's formula shows that  $n^n/n!\sim(2\pi n)^{-1/2}\mathrm e^{n}$, hence $\sum\limits_{i=1}^n\log(n/i)=\log(n^n/n!)$ (your post) and
$$\log\left(n^n/n!\right)=n-\frac12\log(2\pi n)+o(1)=n+o(n)=\Theta(n).$$

Edit (without Stirling's formula)

Call $x_n=\sum\limits_{i=1}^n\log(n/i)=n\log(n)-\sum\limits_{i=1}^n\log(i)$, hence $$
x_{n+1}-x_n=(n+1)\log(n+1)-n\log(n)-\log(n+1)=n\log(1+1/n).
$$
Since $\log(1+u)=u+o(u)$ when $u\to0$, $x_{n+1}-x_n=1+o(1)$ when $n\to\infty$. This implies that $x_n=n+o(n)=\Theta(n)$.