## University Calculus: Early Transcendentals (3rd Edition)

$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{(-1)^{n+1}}{2n-1}$ This implies $\lim\limits_{n \to \infty} \dfrac{\dfrac{(-1)^{n+1}}{2n-1}}{\dfrac{2n}{n}-\dfrac{1}{n}}=\dfrac{0}{2}=0$ Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.