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?

Note that

$$\frac{1}{9n^2 + 3n - 2} = \frac{1}{(3n + 2)(3n - 1)} = \frac{1}{3} \left( \frac{1}{3n - 1} - \frac{1}{3n + 2} \right)$$

for all natural numbers $n$.

The partial sums of the sum that you are interested in are then given by

$$\sum_{n=1}^N \frac{1}{9n^2 + 3n - 2} = \frac{1}{3} \sum_{n=1}^N \left( \frac{1}{3n - 1} - \frac{1}{3n + 2} \right) = \frac{1}{3} \sum_{n=1}^N \left( \frac{1}{3n - 1} - \frac{1}{3(n + 1) - 1} \right)$$

One sees that the sum telescopes; that is to say all of the terms cancel except for the first and last, leaving us with

$$\frac{1}{3} \left( \frac{1}{3 \cdot 1 - 1} - \frac{1}{3(N + 1) - 1} \right) = \frac{1}{6} - \frac{1}{9N + 6}$$

As $N$ tends to $\infty$, this tends to $\frac{1}{6}$, and so we see that the sum converges and that its value is $\frac{1}{6}$.

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.

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