Answer
$\lim\limits_{n \to \infty} a_n=- 5 $ and {$a_n$} is convergent.
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{1-5n^4}{n^4+8n^3}$
$= \dfrac{\lim\limits_{n \to \infty} (\dfrac{1}{n^4}-5)}{\lim\limits_{n \to \infty} (1+\dfrac{8}{n})}$
Thus, $\dfrac{\lim\limits_{n \to \infty} \dfrac{1}{n^4}- \lim\limits_{n \to \infty} 5}{\lim\limits_{n \to \infty} 1+ \lim\limits_{n \to \infty} \dfrac{8}{n}}=\dfrac{0-5}{1+0}$
or, $=- 5$
Thus, $\lim\limits_{n \to \infty} a_n=- 5 $ and {$a_n$} is convergent.
