Answer
$\lim\limits_{n \to \infty} a_n=- \infty $ and {$a_n$} is divergent.
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{2n+1}{1-3 \sqrt n}$
and $\lim\limits_{n \to \infty} \dfrac{2n+1}{1-3 \sqrt n} =\dfrac{\infty}{\infty}$
We need to apply L-Hospital's rule:
$=\lim\limits_{n \to \infty} \dfrac{2}{\dfrac{-3}{2 \sqrt n}} $
or, $= \lim\limits_{n \to \infty} \dfrac{4 \sqrt n}{-3} $
or, $=- \infty$
Thus, $\lim\limits_{n \to \infty} a_n=- \infty $ and {$a_n$} is divergent.
