Not necessarily. Consider the series \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n\) and \(\displaystyle\sum_{n=1}^\infty\frac1n\). Here \(a_n=\dfrac1n\).
The first series converges by the alternating series test, which says \(\sum(-1)^na_n\) converges if \(\left|(-1)^na_n\right|=|a_n|\) is a decreasing sequence and converges to 0. This is the case, as \(a_n=\dfrac1n\to0\) as \(n\to\infty\), and each term is decreasing. (Indeed the series converges to \(-\ln2\).)
On the other hand, the second series is a classic example of a divergent sum.
