\[C=\lim_{n\to\infty}\left(\sum_{k=1}^{n}k\ln k-\left(\tfrac{1}{2}n^{2}+\tfrac{1%
}{2}n+\tfrac{1}{12}\right)\ln n+\tfrac{1}{4}n^{2}\right)=\frac{\gamma+\ln\left%
(2\pi\right)}{12}-\frac{\zeta'\left(2\right)}{2\pi^{2}}=\frac{1}{12}-\zeta'%
\left(-1\right),\]