Answer
$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{(-1)^{n+1}}{2n-1}$
Then $\lim\limits_{n \to \infty}
\dfrac{(-1)^{n+1}}{2n-1}=\lim\limits_{n \to \infty} \dfrac{\dfrac{(-1)^{n+1}}{2n-1}}{\dfrac{2n}{n}-\dfrac{1}{n}}=\dfrac{0}{2}$
or, $=0$
Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.
