2. Let $a_n=\frac1n,$ then obviously $\lim_{n\to\infty}\frac1n=0.$ On the other hand, $a_n$ is a strictly decreasing sequence for $n\ge1.$ Hence, by Leibniz test the series converges.
But, this series converges conditionally, since $\left| a_{n} \right|=\frac{1}{n}$ and $\sum\limits_{n=1}^{\infty }{\frac{1}{n}}$ diverges.