Answer
$\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{n+3}{n^2+5n+6}$
or, $\lim\limits_{n \to \infty}
\dfrac{n+3}{n^2+5n+6}= \lim\limits_{n \to \infty} \dfrac{n+3}{(n+3)(n+2)}$
Now, apply the Sandwich Theorem:
$0<\dfrac{1}{n+2}<\dfrac{1}{n}$
$0\leq \lim\limits_{n \to \infty} \dfrac{1}{n+2}\leq\lim\limits_{n \to \infty} \dfrac{1}{n}$
But $\lim\limits_{n \to \infty} \dfrac{1}{n}=0$, thus we get
$\lim\limits_{n \to \infty} \dfrac{1}{n+2}=0$
Hence, $\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
