Given the series $\sum_{n=1}^{\infty} a_n$ converges, prove $\sum_{n=1}^{\infty}max\{a_n,\frac{1}{n^2}\}$ converges.

Question

I'm given that $\sum_{n=1}^{\infty}a_n$ is a convergent series and I've to prove the series mentioned above converges.
I started by denoting $c_n = max\{a_n,\frac{1}{n^2}\}$.Let $\epsilon >0$
$\sum_{n=1}^{\infty}a_n$ converges $\Longrightarrow$ $\lim_{n\to\infty}a_n = 0 \Longrightarrow \exists N_1\in \mathbb{N}$ s.t $\forall n\ge N_1.$ $|a_n|<\epsilon$
$\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges $\Longrightarrow$ $\lim_{n\to\infty}\frac{1}{n^2} = 0 \Longrightarrow \exists N_2\in \mathbb{N}$ s.t $\forall n\ge N_2.$ $|\frac{1}{n^2}|<\epsilon$.
Taking the index $N=max\{N_1,N_2\}$ will yield that $\forall n>N.$ $|c_n|<\epsilon$.
This implies $lim_{n\to\infty}c_n=0$ which is a good point to start with but I didn't know how to continue from here..
I don't have any information about the sequence $a_n$ and thus I can't use any tests that applies for positive series' and since $|c_n|\ge |\frac{1}{n^2}|$ I can't really bound the desired series by a convergent one.

Answer 1

There's a counterexample. Put $a_n = \frac{(-1)^n}{n}$ and $b_n = max\{a_n,\frac{1}{n^2}\}$. Hence $b_n \ge 0$ and $b_n$ is big if $n$ is even.