## Thomas' Calculus 13th Edition

$\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} is convergent.
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{n+1}{2n})(1-\dfrac{1}{n})$ and $\lim\limits_{n \to \infty} (\dfrac{n+1}{2n})(1-\dfrac{1}{n})=\lim\limits_{n \to \infty} (\dfrac{n+1}{2n}-\dfrac{n+1}{2n^2})$ Also, $\lim\limits_{n \to \infty} (\dfrac{1}{2}+\dfrac{1}{2n})-\lim\limits_{n \to \infty} (\dfrac{1}{2n}+\dfrac{1}{2n^2})=\dfrac{1}{2}$ Hence, $\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} is convergent.