Example 1. Investigate the convergence of the series $\sum_{n=2}^{\infty} \frac{18}{n^{2}+n-2}$ and, if possible, find its sum.
Solution. We have $\frac{18}{n^{2}+n-2}=\frac{6}{n-1}-\frac{6}{n+2}$. Let's form the partial sum and find its limit. Notice which terms cancel each other out (!). We have:

$$
\begin{aligned}
& S_{n}=u_{2}+u_{3}+\ldots+u_{n}= \\
& =6\left[\left(\frac{1}{1}-\frac{1}{A}\right)+\left(\frac{1}{2}-\frac{1}{b}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{A}-\frac{1}{7}\right)+\ldots\right. \\
& +\left(\frac{1}{n-4}-\frac{1 /}{n+1}\right)+\left(\frac{1 /}{n+3}-\frac{1}{n}\right)+ \\
& \left.+\left(\frac{1 /}{n-2}-\frac{1}{n+1}\right)+\left(\frac{1}{n-1}+\frac{1}{n+2}\right)\right]= \\
& =6\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\right)
\end{aligned}
$$

ll

$$
\lim _{n \rightarrow \infty} S_{n}=6 \lim _{n \rightarrow \infty}\left(\frac{11}{6}-\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\right)=11
$$

Answer. The given series converges, and its sum is $S=11$.