Answer
$\lim\limits_{n \to \infty} a_n=- \infty $ and {$a_n$} is divergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{2n+1}{1-3 \sqrt n}$
Check the indeterminate form of limit.
This implies $\lim\limits_{n \to \infty} \dfrac{2n+1}{1-3 \sqrt n} =\dfrac{\infty}{\infty}$
We will have to apply L'Hôpital's rule.
we have $\lim\limits_{n \to \infty} \dfrac{2}{(\dfrac{-3}{2 \sqrt n})}= \lim\limits_{n \to \infty} \dfrac{4 \sqrt n}{(-3)}=- \infty$
Hence, $\lim\limits_{n \to \infty} a_n=- \infty $ and {$a_n$} is divergent.
