I have to solve find the value of $\sum_{k=1}^{n/2} k\log k$ as a part of question.
How should I proceed on this ?
I have to solve find the value of $\sum_{k=1}^{n/2} k\log k$ as a part of question.
How should I proceed on this ?
Got it. The constant in Moron's answer is $C = \log A$, where $A$ is the Glaisher-Kinkelin constant. Thus C = \frac{1}{12} - \zeta'(-1).
The expression $H(n) = \prod_{k=1}^n k^k$ is called the hyperfactorial, and it has the known asymptotic expansion
$H(n) = A e^{-n^2/4} n^{n(n+1)/2+1/12} \left(1 + \frac{1}{720n^2} - \frac{1433}{7257600n^4} + \cdots \right).$ Taking logs and using the fact that $\log (1 + x) = O(x)$ yields an asymptotic expression for the OP's sum $\sum_{k=1}^n k \log k = C - \frac{n^2}{4} + \frac{n(n+1)}{2} \log n + \frac{\log n}{12} + O \left(\frac{1}{n^2}\right),$ the same as the one Aryabhata obtained with Euler-Maclaurin summation.
Here is an asymptotic expression using EulerMcLaurin Summation.
\sum _{k=1}^{n} k \log k = \int_{1}^{n} x \log x\ \text{d}x + (n\log n)/2 + C' + (\log n + 1)/12+ \mathcal{O}(1/n^2)
$ = n^2(2 \log n - 1)/4 + (n\log n)/2 + (\log n)/12 + C + \mathcal{O}(1/n^2)$
for some constant $C$.