Logarithmic Series Evaluation

Question

I was trying to generate a direct formula for this series but I am not sure whether it is possible to do so.

$$1\ln(1) + 2\ln(2) + 3\ln(3) + 4\ln(4)+\dots+(n-1)\ln(n-1) + n\ln(n)$$

Answer 1

By partial summation: $$\sum_{k=1}^{n}k\log k = \binom{n+1}{2}\log n-\sum_{k=1}^{n-1}\binom{k+1}{2}\log\left(1+\frac{1}{k}\right)$$ and using the approximation $\log\left(1+\frac{1}{k}\right)=\frac{1}{k}-\frac{1}{2k(k+1)}-\frac{1}{6k(k+1)^2}+O\left(\frac{1}{k^4}\right)$ we get:

$$\sum_{k=1}^{n}k\log k = \frac{6n^2+6n+1}{12}\log n-\frac{n^2}{4}+O(1).$$

In the $O(1)$-term hides the logarithm of the Glaisher-Kinkelin constant.