Answer
convergent
$S=\dfrac{1}{6}$
Work Step by Step
Given
$$ \sum_{k=1}^{\infty} \frac{1}{9 k^{2}+3 k-2} $$
Since by using partial fractions
\begin{aligned}
\sum_{k=1}^{\infty} \frac{1}{9 k^{2}+3 k-2} &=\sum_{k=1}^{\infty} \frac{1}{(3 k+2)(3 k-1)} \\
&=\sum_{k=1}^{\infty} \frac{1}{3}\left(\frac{1}{(3 k-1)}-\frac{1}{(3 k+2)}\right) \\
&=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\frac{1}{9 k-3}-\frac{1}{9 k+6}\right) \\
&=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{9-3}-\frac{1}{9 k+6}\right)+\left(\frac{1}{18-3}-\frac{1}{18+6}\right)+\cdots\right.\\
&\left.=\lim _{n \rightarrow \infty}\left(\frac{1}{6}-\frac{1}{9 n+6}\right)+\cdots+\left(\frac{1}{9 n-3}-\frac{1}{9 n+6}\right)\right] \\
&=\lim _{n \rightarrow \infty}\left(\frac{1}{6}-\frac{1}{9 n+6}\right) \\
&=\frac{1}{6}-0 \\
&=\frac{1}{6}
\end{aligned}
The series is convergent and the sum is $S=\dfrac{1}{6}$
