Answer
$\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} is convergent
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
(\dfrac{n+1}{2n})(1-\dfrac{1}{n})$
$=\lim\limits_{n \to \infty} (\dfrac{n+1}{2n}-\dfrac{n+1}{2n^2})$
$=\lim\limits_{n \to \infty} (\dfrac{1}{2}+\dfrac{1}{2n})-\lim\limits_{n \to \infty} (\dfrac{1}{2n}+\dfrac{1}{2n^2})$
or, $=\dfrac{1}{2}$
Thus, $\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} is convergent.
