Answer
Convergent with limit $8$
Work Step by Step
Given: $f(x)=\lim\limits_{n\to \infty} \dfrac{24}{n^3}[\dfrac{n(n+1)(2n+1)}{6}]= \lim\limits_{n\to \infty} \dfrac{8n^2+12n+4}{n^2}$
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
$\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty} \dfrac{8+(12/n)+(4/n^2)}{n^2}$
This gives:
$\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty}(8)+\lim\limits_{n\to \infty}\dfrac{12}{n}+\lim\limits_{n\to \infty}\dfrac{4}{n^2}$
Thus, $a_n=8$
Hence, the sequence is convergent with limit $8$
