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