# $\sum a_{n}$ 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?

• The implication you said after "because" is true but it's still possible that $\sum_n a_n$ converges but not absolutely and $\sum_n a_n^2$ does not converge (absolutely nor by any means).
– user562983
Commented Nov 4, 2018 at 23:05
• @max_zorn That's if $\sum a_n$ converges absolutely. Commented Nov 4, 2018 at 23:09
• Yes, you are right Robert, sorry! Commented Nov 4, 2018 at 23:12

The alternating series test gives a wealth of examples. Take $$a_n=(-1)^n/\sqrt{n}$$ for example.

• You beat me to it by $14$ seconds. Commented Nov 4, 2018 at 23:07
• Are there any that don't involve a $(-1)^n$ term? Commented Nov 5, 2018 at 4:27
• @Ingolifs: $\sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds. Commented Nov 5, 2018 at 8:00

Try $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$$

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.

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 $$f:\mathbb R\rightarrow \mathbb R$$,

$$\sum a_n ~CV \Rightarrow \sum f(a_n) ~CV \quad \text{iff} \quad \exists \eta>0,\exists \lambda\in \mathbb R,\forall x\in ]-\eta,\eta[, \quad f(x)=\lambda x$$

$$\sum a_n ~CV \Rightarrow \sum f(a_n) ~ACV \quad \text{iff} \quad \exists \eta>0,\forall x\in ]-\eta,\eta[, \quad f(x)=0 \quad\quad\quad~~$$

$$\sum a_n ~ACV \Rightarrow \sum f(a_n) ~ACV \quad \text{iff} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ $$\quad f(0)=0 \text{ and }\exists \eta>0,\exists M >0,\forall x\in ]-\eta,\eta[, |f(x)|\leq M |x|$$

$$\sum a_n ACV \Rightarrow \sum f(a_n) ~CV \quad \text{iff} \quad \sum a_n ~ACV \Rightarrow \sum f(a_n) ~ACV~~~~~~~~~~~~~~$$

• $\sum_n a_n^2$ converges for this one. Commented Nov 4, 2018 at 23:07
• oups I answer too fast ^^ Commented Nov 4, 2018 at 23:10
• I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements. Commented Nov 5, 2018 at 1:23
• While I realize it's a standard notation, $x\in]-\eta,\eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $x\in(-\eta,\eta)$ is much better, though $|x| < \eta$ is better than both. Commented Nov 5, 2018 at 2:33
• I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $\sum a_n$ converges then $\sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$." Commented Nov 5, 2018 at 2:34