Prove/disprove: Let $\{a_n\}_{n=1}^\infty$. If $\sum_{i=0}^\infty a_n$ converges $\implies$ $\sum_{i=0}^\infty \max\{a_n, {1 \over n^2}\}$ converges.

Question

So I think this correct and have made an attempt to prove it and I would like to know if it is okay:

Let $\{c_n\}_{n=1}^\infty$ s.t $\forall n \in \mathbb {N}: \; c_n = \max\{a_n, {1 \over n^2}\}$.

Notice $\sum_{i=0}^\infty c_n \leq \sum_{i=0}^\infty (a_n+{1 \over n^2}) \leq \sum_{i=0}^\infty a_n + \sum_{i=0}^\infty {1 \over n^2}$.

From arthmetics of columns we have $\sum_{i=0}^\infty (a_n+{1 \over n^2})$ converges.

Therefore we have $\sum_{i=0}^\infty c_n$ converges as needed.

Edit:

So as explained in the comments my mistake was assuming $\sum_{i=0}^\infty a_n$ converges $\iff \sum_{i=0}^\infty a_n$ converges absolutely. As written in the comment section by @Sangchul Lee, using $a_n = {(-1)^n \over n}$ would disprove the theorem.

Thanks to all!