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

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 $\sum Aa_n=A\sum a_n$.
