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?

Use partial sums:
$$
\frac{1}{9n^2+3n-2} = \frac{1}{(3n-1)(3n+2)} = \frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)
$$
The sum you need is a telescopic sum, I hope you can work out the rest.
