Answer
$\dfrac{3}{2}$
Work Step by Step
Consider $s_n=\dfrac{9}{(3n-1)(3n+2)}$
Re-write the given series as: $s_n=\dfrac{9}{(3n-1)(3n+2)}=\dfrac{3}{(3n-1))}-\dfrac{9}{(3n+2)}$
$s_n=(\dfrac{3}{2}-\dfrac{3}{5})+(\dfrac{3}{5}-\dfrac{3}{8})+(\dfrac{3}{8}-\dfrac{3}{11})+(\dfrac{3}{11}-\dfrac{3}{14})...+(\dfrac{3}{3n-1}-\dfrac{3}{3n+2})$
or,
$s_n=\dfrac{3}{2}-\dfrac{3}{3n+2}$
Apply limits, we get:
$\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(\dfrac{3}{2}-\dfrac{3}{3n+2})$
After simplifications, we get
$\lim\limits_{n \to \infty}s_n=(\dfrac{3}{2}-0)$
Hence, $\lim\limits_{n \to \infty}s_n=\dfrac{3}{2}$
