I am familiar with the proof using the comparison test, but I was wondering if it is possible to prove this result in the following way:
We know that the sequence $a_n$ converges, therefore $\exists M$ such that $M \geq a_n, \forall n$. So,
$$|a_n^2 + a_{n+1}^2 + \ldots + a_{n+p}^2| \leq M |a_n + a_{n+1} + \ldots + a_{n+p}|$$
Now, given that $\sum a_n$ is convergent, by Cauchy's criterion, $\forall \epsilon > 0, \exists n_0$ such that $\forall n > n_0, p \in \mathbb{N}$ we have
$$|a_n + a_{n+1} + \ldots + a_{n+p}| < \frac{\epsilon}{M}$$
Thus,
$$|a_n^2 + a_{n+1}^2 + \ldots + a_{n+p}^2| < M \frac{\epsilon}{M} = \epsilon $$
Best Answer
We can see that $\displaystyle\lim_{n\rightarrow \infty} \frac{a_{n}^{2}}{a_n}=\lim_{n\rightarrow \infty} a_n=0$.
Based on this,we can know that the original proposition holds.