Is it true that $\displaystyle \sum_{n=1}^{\infty} {a_{n}}^2 $ is convergent if $\displaystyle \sum_{n=1}^{\infty} {a_{n}} $ is convergent? Justify your answer.
Any hints?
Remember that if $\displaystyle \vert a \vert \leq 1$ then $\displaystyle a^2 \leq \vert a \vert$. Now also remember if the series is convergent what is $\displaystyle \lim_{n \to \infty} \ a_n $?
Although now that I think of it, I'm not sure it works in general if the series does not converge absolutely...if it does, following the above idea suffices, but if the series converges only conditionally I'm not so sure.
Yes, by the comparison test.
Since $\displaystyle \lim_{n\to\infty}a_n=0$ then there exists $\displaystyle N$ such that $\displaystyle |a_n|<1\Longrightarrow a_n^2<|a_n|$ for all $\displaystyle n>N$. Then apply the comparison test, but you can assume that $\displaystyle a_n\ge 0$.
EDIT:** Sorry but the answer is NO. Take $\displaystyle a_n=\frac{(-1)^n}{\sqrt{n}}$ **
But there's a problem, if for instance $\displaystyle \{a_n \}$ alternates sign, comparison test no longer applies since $\displaystyle \{a_n^2 \}$ is always positive, but $\displaystyle \{a_n \}$ isn't. Here's an example if $\displaystyle \{a_n \}= \{\frac{(-1)^n}{n} \}$ then $\displaystyle 1-\frac{1}{2} \ngeq 1+\frac{1}{4}$
Just thought i'd add in this one point...
If $\displaystyle \sum_n a_n$ is a convergent series of POSITIVE TERMS then $\displaystyle \sum_n a_n^2$ converges...
Proof,
$\displaystyle a_n$ must be bounded, i.e. for some $\displaystyle M > 0$ we have an $\displaystyle a_n \leq M$ for all n. Then $\displaystyle a_n^2 \leq Ma_n$ so by comparison test $\displaystyle a_n^2$ converges.