Answer
$\lim\limits_{n \to \infty} a_n=-1 $ and {$a_n$} is convergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{1-2n}{1+2n}$
Then, $ \lim\limits_{n \to \infty}
\dfrac{1-2n}{1+2n}= \dfrac{\lim\limits_{n \to \infty} (\dfrac{1}{n}-2)}{\lim\limits_{n \to \infty} (\dfrac{1}{n}+2)}$
This implies, we have
$\dfrac{\lim\limits_{n \to \infty} \dfrac{1}{n}- \lim\limits_{n \to \infty} 2}{\lim\limits_{n \to \infty} \dfrac{1}{n}+\lim\limits_{n \to \infty} 2}=\dfrac{0-2}{0+2}=-1$
Hence, $\lim\limits_{n \to \infty} a_n=-1 $ and {$a_n$} is convergent.
