Prove if the sum is convergent, then the square is convergent

Question

I've seen multiple ways on how to solve it online (and most likely the majority are wrong), but I don't know how to solve this fully so that I could get the full grade during my exam.

The question is as follows:

Prove that if $\sum a_n$ is convergent with $a_n > 0$ for all $n$, then $\sum a^2_n$ is also convergent.

I understand that the starting point to prove this is that $\lim a_n =0$, but after that I really don't know what I'm supposed to say.

Answer 1

Since $\lim_{n\to\infty}a_n=0$, you have $a_n\leqslant1$ if $n$ is large enough. But then ${a_n}^2\leqslant a_n$ if $n$ is large enough and therefore you can apply the comparison test.

Answer 2

For any $\epsilon >0$, there exists an $N$ such that all $n>N$ satisfy $|a_n|<\epsilon$. Hence $|a_n|^2<\epsilon |a_n|$.

Answer 3

Then for large $n$, $a_{n}<1$, then $a_{n}^{2}\leq a_{n}$ for all such $n$ and hence $\displaystyle\sum_{n\geq N}a_{n}^{2}\leq\sum_{n\geq N}a_{n}<\infty$.