Can you help me with this problem related to absolute convergence [closed]

if {$a_{n}$} is a real sequence such that lim $n^{2}a_{n}$ exists in R, show that the series $\sum a_{n}$ is absolutely convergent.

All I have in mind is that the sequence $a_{n}$ must be a convergent sequence and somehow use that to prove that the series converges absolutely. I tried using the ratio test but I can't move forward.

Any help will be appreciated as I really need this exam of mine to go well

You need to show that $ \sum_{n=1}^\infty | a_n | = L $ for some $ L \ge 0 $. Since $ \sum_{n=1}^{r} | a_n | \le \sum_{n=1}^{r + s} | a_n | $ for all $ s \ge 0 $, it is enough to show that, given any $ \epsilon > 0 $, there is $ N $ such that for all $ s \ge 0 $, $ \sum_{n = N}^{N + s} | a_n | < \epsilon $. Now, suppose $ \lim_{n \to \infty} n^2 a_n = c $. Then, there is $ N_1 $ such that for all $ n \ge N_1 $, $ | a_n | \le \frac{(|c| + 1)}{n^2} $. Since the series $ \sum_{n=1}^\infty \frac{1}{n^2} $ is convergent, given any $ \epsilon > 0 $, there is $ N_2 $ such that $ \sum_{n=N_2}^{N_2 + s} \frac{1}{n^2} < \frac{\epsilon}{|c| + 1} $ for all $ s \ge 0 $. Taking $ N = \max\{ N_1, N_2 \} $, you have $ \sum_{n = N}^{N + s} | a_n | \le (|c| + 1) \sum_{n = N}^{N + s} \frac{1}{n^2} < (|c| + 1) \frac{\epsilon}{|c| + 1} = \epsilon $, which completes the proof.