$\sum a_{n}$ converges but $\sum a_{n}^2 $ diverges? I have to give an example of a convergent series $\sum a_{n}$ for which $\sum a_{n}^2 $ diverges.
I think that such a series cannot exist because if $\sum a_{n}$ converges absolutely then $\sum a_{n}^2 $ will always converge right?
 A: More generally,
if $a_n = \dfrac{(-1)^n}{n^{1/(2m)}}$,
then
$\sum_{n=1}^{\infty} a_n
$
converges for integer $m \ge 0$
and
$\sum_{n=1}^{\infty} a_n^{2m}
$
diverges.
A: The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/\sqrt{n}
$$
for example.
A: Try $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$$
A: The simplest example I have in mind is : 
$$\sum_{n=1}^{+\infty}\frac{(-1)^n}{\sqrt{n}} $$
Beware that convergent (CV) and absolutely convergent (ACV) can be very different. 
Indeed, if $\sum_{n=1}^{+\infty} a_n $ is ACV then, $\sum_{n=1}^{+\infty} a_n^2$ is also ACV. To prove it, you can simply notice that, since $\lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$. 
You can also ask yourself a more general question : 
What are the function $f:\mathbb R \rightarrow \mathbb R$ such that for all 
$\sum a_n$ CV (resp. ACV), $\sum f(a_n)$ is CV (resp. ACV).
It is a (difficult) exercise to show that for $f:\mathbb R\rightarrow \mathbb R$,
$$\sum a_n ~CV \Rightarrow \sum f(a_n) ~CV \quad \text{iff} \quad \exists \eta>0,\exists \lambda\in \mathbb R,\forall x\in ]-\eta,\eta[, \quad f(x)=\lambda x $$
$$\sum a_n ~CV \Rightarrow \sum f(a_n) ~ACV \quad \text{iff} \quad \exists \eta>0,\forall x\in ]-\eta,\eta[, \quad f(x)=0 \quad\quad\quad~~$$
$$\sum a_n ~ACV \Rightarrow \sum f(a_n) ~ACV \quad \text{iff} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ $$\quad f(0)=0 \text{ and }\exists \eta>0,\exists M >0,\forall x\in ]-\eta,\eta[, |f(x)|\leq M |x| $$
$$\sum a_n ACV \Rightarrow \sum f(a_n) ~CV \quad \text{iff} \quad \sum a_n ~ACV \Rightarrow \sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$
