Check convergence and find the sum $\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}$ 
$$\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}$$

I have starting an overview about series, the book starts with geometric series and emphasizing that for each series there is a corresponding infinite sequence.
For convergence I can look at the partial sums, but how can I find the sum of the series?

It is a telescopic series:
$$\sum_{n=1}^{N} \frac{1}{9n^2+3n-2}=\sum_{n=1}^{N}\frac{1}{(3n+2)(3n-1)}=\sum_{n=1}^{N}\left(\frac{1/3}{3n-1}-\frac{1/3}{3n+2}\right)\\=\sum_{n=1}^{N}\frac{1/3}{3n-1}-\sum_{m=2}^{N+1}\frac{1/3}{3m-1}=\frac{1/3}{3\cdot 1-1}-\frac{1/3}{3(N+1)-1}\to \frac{1}{6}\quad \mbox{as $N\to+\infty$}.$$
