How to prove this inequality? $\sum_{k=2}^{n-1}k\log k < \frac{1}{2} n^2\log n-\frac{1}{8}n^2$?

Question

How can I prove this inequality?

$$\sum_{k=2}^{n-1}k\log k < \frac{1}{2} n^2\log n-\frac{1}{8}n^2.$$

Answer 1

Summation by parts is enough. We have:

$$ \sum_{k=1}^{n-1} k\log k = \frac{n^2-n}{2}\cdot\log n-\sum_{k=1}^{n-2}\frac{k^2-k}{2}\cdot\log\left(1+\frac{1}{k}\right)$$ but $k\mapsto k\cdot \log\left(1+\frac{1}{k}\right)$ is an increasing function, hence:

$$ \sum_{k=2}^{n-1} k\log k \leq \frac{n^2-n}{2}\cdot \log n- \sum_{k=2}^{n-2}(k-1)\log\left(\frac{3}{2}\right) $$ leads to the stronger bound:

$$ \sum_{k=2}^{n-1} k\log k \leq \frac{n^2-n}{2}\cdot \log n - \frac{(n-2)(n-3)}{2}\cdot\log\left(\frac{3}{2}\right) $$ for any $n\geq 4$.