$\sum_\limits{n=0}^{\infty} a_n$ converges $\implies \sum_\limits{n=0}^{\infty} a_n^2$ converges 
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?

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.
