$\sum_\limits{n=0}^{\infty} a_n$ converges $\implies \sum_\limits{n=0}^{\infty} a_n^2$ converges [duplicate]

Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.

This is in the section on the Comparison Test so that must be what I'm supposed to use. But I don't see how. $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$. And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges. Hmm. Any hints?

If we know that $a_n\to 0$ then it is going to satisfy $0\le a_n<1$ after a while, that is $a_n^2\le a_n$ for $n\ge N$ with some $N>0$. It makes the comparison for the tails valid $$ \sum_{n=N}^\infty a_n^2\le\sum_{n=N}^\infty a_n. $$ Convergence of the tails is all we need for convergence.

Hint: as an alternative way, it's probably worth mentioning that for positive terms: $$\left(\sum_{n=0}^{k} a_n\right)^2 \geq \sum_{n=0}^{k} a_n^2$$ Now, taking the limit $$\left(\sum_{n=0}^{\infty} a_n\right)^2 \geq \sum_{n=0}^{\infty} a_n^2$$

For any $\epsilon>0$, there must be an $N$ such that $a_n<\epsilon$ for all $n>N$. If this were not true, you'd be adding an infinite amount of terms which didn't approach zero, meaning the sum would diverge, which, according to the statement, is false.

Let $\epsilon=1$ so that we have some $N$ such that $a_n<1$ for all $n>N$. It should then be clear that for any $0<a_n<1$, we have $0<(a_n)^2<a_n<1$.

So, we have

$$\sum_{n>N}(a_n)^2<\sum_{n>N}a_n$$

And since $\sum_{n=1}^N(a_n)^2$ is finite, we show by comparison test that

$$\sum_{n\ge0}(a_n)^2<\sum_{n=1}^N(a_n)^2+\sum_{n>N}a_n$$

which converges.

If $a_n\ge0$ for all $n$ and $\sum_{n=1}^\infty a_n=A$, then $A\ge a_n$ for all $n$, which implies $a_n^2\le Aa_n$. So you can compare $\sum a_n^2$ to $\sum Aa_n=A\sum a_n$.