Answer
$\lim\limits_{n \to \infty} a_n=\sqrt 2$ and {$a_n$} is convergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\sqrt{\dfrac{2n}{n+1}}$
$ \lim\limits_{n \to \infty} \sqrt{\dfrac{2n}{n+1}}=\sqrt{ \lim\limits_{n \to \infty}\dfrac{2n}{n+1}}=\sqrt{ \lim\limits_{n \to \infty}\dfrac{2}{1+\dfrac{1}{n}}}$
Thus, $\lim\limits_{n \to \infty} a_n=\sqrt 2$
Hence, $\lim\limits_{n \to \infty} a_n=\sqrt 2$ and {$a_n$} is convergent.
