Asymptotics of $\sum\limits_{i=1}^n i\log i$ I'm practicing complexity analysis for a problem and came across:
$$\sum_{i = 1}^{n} {\log(i!)}$$
I know that $\log(i!) = \Theta(i\log(i))$ so I further simplified to the sum to:
$$\sum_{i = 1}^{n} {i\log(i)}$$
Breaking this out gives me:
$$1\log(1) + 2\log(2) + 3\log(3) ... + n\log(n)$$
Since we have $1 + 2 + 3 + ... + n$ terms I believe there will be $\frac{n(n+1)}{2}$ terms, thus simplifying the sum to:
$$\frac{n(n+1)}{2}\sum_{i = 1}^{n} {\log(i)}$$
and from my research $$\sum_{i = 1}^{n} {\log(i)} = \log(n!)$$
this giving me the sum total:
$$\frac{n(n+1)}{2}\log(n!) = O(n^2\log(n!))$$
However given my other question on SO it appears this calculation is incorrect?
Could someone clear this up?
 A: Your reasoning goes awry when you say "I believe there will be $\frac{n(n+1)}{2}$ terms." (actually even before, as we will see.) You only have $n$ terms, and cannot take the number "outside" the sum. 
The $i$th term is $i\log i$, it is not $\log i$ or anything else. At that point, there is no "simplifying and taking outside the sum" to do — you have to deal with the summands as they are.

Now, for the original question, and more on your approximation: you would have to justify that. Namely, it is very handwavy, and it's better to make it precise.
You have
$$
\sum_{i=1}^n \ln(i!)
= \sum_{i=1}^n \sum_{j=1}^i \ln j
= \sum_{j=1}^n \sum_{i=j}^n \ln j
= \sum_{j=1}^n (n-j+1) \ln j
= (n+1)\sum_{j=1}^n \ln j - \sum_{j=1}^n j\ln j \tag{1}
$$
which is not a first glance the same thing...$^{(\dagger)}$ Let's deal with each term separately.


*

*As we know (see for instance this other question that 
$
\sum\limits_{j=1}^n \ln j = n \ln n + O(n)
$, the first term is 
$$
(n+1)\sum_{j=1}^n \ln j = n^2\ln n + O(n). \tag{2}
$$

*By a comparison series/integral (since $f$ defined by $f(x) = x\ln x$ is monotone, and nice; see again this other question for more details on the method if you are not familiar with it), it is not hard to show that
$$
\sum_{j=1}^n j \ln j = \int_1^n x\ln x dx + O(n) = \frac{1}{2}n^2 \ln n + O(n^2). \tag{3}
$$
Combining (2) and (3) into (1), we get
$$
\sum_{i=1}^n \ln(i!)
= n^2\ln n - \frac{1}{2}n^2 \ln n + O(n^2) = \frac{1}{2}n^2 \ln n + O(n^2).
$$

$(\dagger)$ Actually, your original approximation is true, because of the theorem below. However, you may want to cite, or justify it before making such approximations — if you do not know how or why they hold, there is a very high probability you'll end up making a mistake very often.

Theorem. Let $(a_n)_n, (b_n)_n$ be two non-negative sequences such that $a_n\sim_{n\to\infty} b_n$. Then the series $\sum\limits_n a_n$ converges iff  the series $\sum\limits_n b_n$ converges; moreover:
  
  
*
  
*If the series converge, then $\sum\limits_{n=N}^\infty a_n \sim_{N\to\infty} \sum\limits_{n=N}^\infty b_n$ (the remainders are equivalent);
  
*If the series diverge, then $\sum\limits_{n=1}^N a_n \sim_{N\to\infty} \sum\limits_{n=1}^N b_n$ (the partial sums are equivalent).
  

A: Your estimate is wrong since there is no reason to go from 
$$\sum_{i=1}^n i \log i$$ to
$$\frac{n(n+1)}{2}\sum_{i=1}^n \log i$$
these two functions are not asymptotically equivalent. A good idea is the following: approximate the sum with an integral
$$\sum_{i=1}^n i \log i = \int_1^n x \log x \  \mathrm{d}x + O(n \log n)$$
The integral can be computed exactly by integrating by parts
$$\int_1^n x \log x \  \mathrm{d}x = \frac{1}{2} n^2 \log n -\frac{1}{4} (n^2-1)  $$
so that the final estimate is
$$\sum_{i=1}^n i \log i \sim \frac{1}{2} n^2 \log n = O( n^2 \log n)$$
A: $$\sum_{i=1}^{n}\log(i!)=\sum_{i=1}^{n}\sum_{j=1}^{i}\log(j) = \sum_{j=1}^{n}(n+1-j)\log(j)=\sum_{k=1}^{n}k\log(n+1-k) \tag{1}$$
is straightforward to study through Abel's summation formula:
$$\sum_{k=1}^{n}k\log(n+1-k) = \int_{1}^{n}\frac{\lfloor x\rfloor(\lfloor x\rfloor+1)}{2(n+1-x)}\,dx \tag{2} $$
from which
$$\boxed{ \sum_{i=1}^{n}\log(i!) = \frac{n^2}{2}\log(n)-\frac{3n^2}{4}+O(n\log n)}\tag{3}$$
easily follows.
