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

take two sequences $x_n$=$\sum_{k=0}^{n}a_k$ and $t_n$=$\sum_{k=0}^{n}(a_k)^2$
now $a_n$> 0 for all n$\in$N so we can write $t_n$ < $(x_n)^2$ for all n. now as $\sum a_n$ is convergent as lim$x_n$ exists.. then the sequence $x_n$ is convergent and so it is bounded ..then there exists M>0 s.t $x_n$< M for all n$\in$N . so $t_n$< $M^2$ for all n$\in$N . now $t_n$ is increasing strickty and bounded above by $M^2$
so $t_n$ is convergent..means $\sum (a_n)^2$ is convergent
