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?

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    $\begingroup$ 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). $\endgroup$
    – user562983
    Commented Nov 4, 2018 at 23:05
  • $\begingroup$ @max_zorn That's if $\sum a_n$ converges absolutely. $\endgroup$ Commented Nov 4, 2018 at 23:09
  • $\begingroup$ Yes, you are right Robert, sorry! $\endgroup$
    – max_zorn
    Commented Nov 4, 2018 at 23:12

4 Answers 4


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

  • $\begingroup$ You beat me to it by $14$ seconds. $\endgroup$ Commented Nov 4, 2018 at 23:07
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    $\begingroup$ Are there any that don't involve a $(-1)^n$ term? $\endgroup$
    – Ingolifs
    Commented Nov 5, 2018 at 4:27
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    $\begingroup$ @Ingolifs: $\sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds. $\endgroup$
    – user21820
    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 $f:\mathbb R\rightarrow \mathbb R$,

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

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

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

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

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    $\begingroup$ $\sum_n a_n^2$ converges for this one. $\endgroup$ Commented Nov 4, 2018 at 23:07
  • $\begingroup$ oups I answer too fast ^^ $\endgroup$ Commented Nov 4, 2018 at 23:10
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    $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2018 at 1:23
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    $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2018 at 2:33
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    $\begingroup$ 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 $x=0$." $\endgroup$
    – Winther
    Commented Nov 5, 2018 at 2:34

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