I was asked that, giving $$\sum_{n=1}^\infty a_n,$$ find $a_n$ such that the series diverges, yet
$$a_n\le\frac{1}{2^n}$$
I came up with this: $$\sum_{i=1}^n-\frac{1}{n}$$ My logic is that I know $1/n$ would diverge, and I know it would be less than $1/2^n$ if I put a negative sign. Am I correct? If not, what value for an would solve the problem?
You are correct. If we had that $$|a_n|\le \frac{1}{2^n}$$ then the series would necessarily converge by the comparison test. Thus the only way around this is to have $|a_n|$ large but $a_n$ negative.