Answer
Convergent with limit $\dfrac{3}{2}$
Work Step by Step
Given: $f(x)=\lim\limits_{n\to \infty} \dfrac{3}{n^2}[\dfrac{n(n+1)}{2}]=\lim\limits_{n\to \infty} \dfrac{3n^2+3n}{2n^2}$
The sequence converges when the limit $\lim\limits_{n\to \infty} a_n$ exists and when the limit $\lim\limits_{n\to \infty} a_n$ does not exist, then sequence diverges.
Here, we have
$\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty}\dfrac{3n^2+3n}{2n^2}$
This gives:
$\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty}\dfrac{3+3/n}{2}$
or, $=\dfrac{\lim\limits_{n\to \infty}(3)+\lim\limits_{n\to \infty}(3/n)}{\lim\limits_{n\to \infty} (2)}$
Thus, $a_n=\dfrac{3}{2}$
Hence, the sequence is convergent with limit $\dfrac{3}{2}$