When I attempted this question I thought that it was necessary but I’m not sure why it’s not necessary because if I take for example
$$a_n = \begin{cases}-(2)^n&\text{if $n$ is odd}\\\frac{1}{2}& \text{if $n$ is even}\end{cases}$$
Would the sum of this sequence diverge to minus infinity? So would this show that $a_n$ does not have to be non-positive eventually?
That sequence is a valid counterexample. Another one would be: $$ a_n = \begin{cases} 1 & \text{$n$ is odd} \\ -2 & \text{$n$ is even} \end{cases} $$ Notice that the sum of any two consecutive terms is $-1$. So the partial sum can be expressed in closed form: $$ \sum_{k=1}^n a_n = \begin{cases} -\frac{n}{2} & \text {$n$ is even} \\ -\frac{n}{2} + \frac{3}{2} & \text{$n$ is odd} \end{cases} $$ Then you can show that the partial sums tend to $-\infty$ as $n\to\infty$.