15. Given the function $y=\frac{x^{2}-x+n}{x^{2}+x+1}\left(n \in \mathbf{N}_{+}\right)$ with the minimum value $a_{n}$, the maximum value $b_{n}$, and $c_{n}=\frac{n}{4}\left(1+3 a_{n} b_{n}\right)$.
(1) Find the general term formula for the sequence $\left\{c_{n}\right\}$;
(2) Prove that $\frac{3}{2}-\frac{1}{n+1}<\sum_{k=1}^{n} \frac{1}{c_{k}}<2-\frac{1}{n}(n \geqslant 2)$.
15. (1) Transform the function into $(y-1) x^{2}-(y+1) x+y-n=0$, so $(y+1)^{2}-4(y-1)(y-n) \geqslant 0$. According to the conditions, the maximum and minimum values $a_{n}$ and $b_{n}$ are the roots of the equation $(y+1)^{2}-4(y-1)(y-n)=0$. By Vieta's formulas, we have $a_{n} b_{n}=\frac{4 n-1}{3}$. Therefore, $C_{n}=n^{2}$, which means the general term formula is $C_{n}=n^{2}\left(n \in \mathbf{N}_{+}\right)$;
(2) First, we can prove that when $n \geqslant 2$, $\frac{1}{n(n+1)}<\frac{1}{n^{2}}<\frac{1}{(n-1) n}$. And $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, $\frac{1}{(n-1) n}=\frac{1}{n-1}-\frac{1}{n}$, so we have $\sum_{k=2}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)<\sum_{k=2}^{n} \frac{1}{C_{k}}<\sum_{k=2}^{1}\left(\frac{1}{k-1}-\frac{1}{k}\right)$. And $\sum_{k=2}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{1}{2}-\frac{1}{n+1}, \sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{1}{n}$, hence $\frac{3}{2}-\frac{1}{n+1}<\sum_{k=1}^{n} \frac{1}{C_{k}}<2-\frac{1}{n}$.