I want to ask something about:
"Since $i \log_e i$ is concave upwards, it is easy to show that $$\sum_{i=2}^{n-1} i \log_e i \leq \int_2^n x \log_e x \,dx \leq \frac{n^2 \log_e n}{2}-\frac{n^2}{4}$$"
$i \log_e i$ is concave upwards because the second derivative of $i \log_e i$ is positive. Correct?
Can you explain the inquality? I don't understand how we show it?
On
Hint1: (for left inequality)
$f(x)=x\ln x$ is increasing on $[2,\infty)$ and
$$\int_2^nf(x)\,dx = \sum_{i=2}^{n-1} \int_i^{i+1}f(x)\,dx \geq \sum_{i=2}^{n-1} f(i).$$
Hint2: (for right inequality) $$\int x\ln x \,dx = F(x) + C$$
where $$ F(x)= \frac{1}{2}x^2\ln x-\frac{1}{4}x^2,$$
and $$F(2) >0.$$
Convexity is not really needed, we can just use partial summation and the inequality $\frac{2}{2n+1}\leq\log\left(1+\frac{1}{n}\right)\leq\frac{1}{n}$:
$$\sum_{n=1}^{N-1}n\log n = \frac{N(N-1)}{2}\log N-\sum_{n=1}^{N-2}\frac{n(n+1)}{2}\log\left(1+\frac{1}{n}\right),$$
$$\sum_{n=1}^{N-2}\frac{n(n+1)}{2}\log\left(1+\frac{1}{n}\right)\leq\sum_{n=1}^{N-2}\frac{n+1}{2}=\frac{N^2-N-2}{4},$$
$$\sum_{n=1}^{N-2}\frac{n(n+1)}{2}\log\left(1+\frac{1}{n}\right)\geq\sum_{n=1}^{N-2}\frac{n^2+n}{2n+1}\\=\frac{1}{4}\sum_{n=1}^{N-2}\left(2n+1-\frac{1}{2n+1}\right)\geq\frac{3N^2-7N+2}{12}.$$