[Math Problem 294.4] For $n=0,1,2, \cdots$ find the sum:
$$S_{n}=\sum_{k=0}^{n} \operatorname{arccot}\left(k^{2}-k+4+\frac{6}{k^{2}-k-1}\right)$$

and find the value of $\lim S_{n}$.
Solution: It is easy to see that $S_{0}=\operatorname{arccot}(-2)=\pi-\operatorname{arccot} 2, S_{1}=$ $2 \operatorname{arccot}(-2)=2 \pi-2 \operatorname{arccot} 2$. For $n=2,3, \cdots$, it is easy to see that
$$\left.S_{n}=S_{1}+\sum_{k=2}^{n} \arctan \frac{A_{k}-1}{\left(A_{k}+1\right.}\right)\left(A_{k}+2\right),$$

where $A_{k}=k(k-1), k=0,1,2, \cdots$. Let $B_{k}=\frac{A_{k}+1}{A_{k+1}+1}$, $k=0,1,2, \cdots$. We will prove that for $k=2,3, \cdots$, the following holds: $\arctan \frac{A_{k}-1}{\left(A_{k}+1\right)\left(A_{k}+2\right)}=\arctan B_{k}-\arctan B_{k-1}$. Clearly, for $k=2,3, \cdots$, $B_{k}>B_{k-1}>0$, and it is easy to calculate that
$$\begin{aligned}
& \frac{B_{k}-B_{k-1}}{1+B_{k} B_{k-1}}=\frac{\left(A_{k}+1\right)^{2}-\left(A_{k-1}+1\right)\left(A_{k+1}+1\right)}{\left(A_{k}+1\right)\left(A_{k+1}+A_{k-1}+2\right)} \\
= & \frac{k^{2}-k-1}{\left(k^{2}-k+1\right)\left(k^{2}-k+2\right)}=\frac{A_{k}-1}{\left(A_{k}+1\right)\left(A_{k}+2\right)}
\end{aligned}$$

Thus, (*) holds, and for $n=2,3, \cdots$, we have
$$\begin{aligned}
S_{n} & =S_{1}+\sum_{k=2}^{n}\left(\arctan B_{k}-\arctan B_{k-1}\right) \\
& =S_{1}+\arctan B_{n}-\arctan B_{1} \\
& =2 \pi-2 \operatorname{arccot} 2+\arctan \left(\frac{n^{2}-n+1}{n^{2}+n+1}\right)-\arctan \frac{1}{3} \\
& =2 \pi-2 \arctan 3+\arctan \left(\frac{n^{2}-n+1}{n^{2}+n+1}\right)
\end{aligned}$$

Therefore, we have
$$\begin{aligned}
S_{n}= & \left\{\begin{array}{l}
\pi-\operatorname{arccot} 2, n=0 ; \\
2 \pi-2 \operatorname{arccot} 2, n=1 ; \\
2 \pi-\arctan 3+\arctan \left(\frac{n^{2}-n+1}{n^{2}+n+1}\right), n \geqslant 2 .
\end{array}\right. \\
\text { Hence } \lim _{n \rightarrow \infty} S_{n} & =2 \pi-\arctan 3+\arctan \left(\lim _{n \rightarrow \infty} \frac{n^{2}-n+1}{n^{2}+n+1}\right) \\
& =2 \pi-\arctan 3+\arctan 1=\frac{9}{4} \pi-\arctan 3 .
\end{aligned}$$