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I'm trying to re-learn my undergrad math, and I'm using Stephen Abbot's Understanding Analysis. In section 2.7, he has the following exercise:

Exercise 2.7.5 (a) Show that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ also converges absolutely. Does this proposition hold without absolute convergence?

I'm posting about this here because my answer to that last question about whether the proposition holds without absolute convergence is "Yes", but my suspicions are raised by him merely asking the question. Usually questions like this are asked to point out that certain conditions are necessary in the statement of propositions, theorems, etc. I just want to see if I'm missing something here.

Anyway, here's how I prove the absolute convergence of $\sum{a_n^2}$, and note that I never use the fact that $\sum{a_n}$ is absolutely convergent:

Proof: Let $s_n = \sum_{i=1}^n{a_n^2}$. I want to show that $(s_n)$ is a Cauchy sequence. So, let $\epsilon > 0$ and $n > m$, and consider, \begin{equation} \begin{aligned} |s_n - s_m| &= |(a_1^2 + a_2^2 + \cdots a_n^2) - (a_1^2 + a_2^2 + \cdots a_m^2)|\\ &= |(a_{m+1})^2 + (a_{m+2})^2 + \cdots + a_n^2|\\ &= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\ \end{aligned} \end{equation} Now, since $\sum{a_m}$ converges, the sequence $(a_m)$ has limit 0. Therefore I can choose an $N$ such that $|a_m| < \sqrt{\epsilon/n}$ for all $m > N$. So for all $n > m> N$, we have, \begin{equation} \begin{aligned} |s_n - s_m| &= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\ &< \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n}}\right)^2\\ &< \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2\\ &= \epsilon \end{aligned} \end{equation} Therefore $(s_n)$ is Cauchy and $\sum{a_n^2}$ converges. And since $\sum{a_n^2} = \sum{|a_n^2|}$, the series $\sum{a_n^2}$ is absolutely convergent. ∎

Am I doing something wrong here? Am I right in thinking that $\sum{a_n}$ need not be absolutely convergent for $\sum{a_n^2}$ to be absolutely convergent?

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    $\begingroup$ $\sum a_n$ for $a_n=(-1)^n/\sqrt{n}$ converges (non-absolutely); with absolute convergence just note that $a_n^2<|a_n|$ for $n$ big enough $\endgroup$
    – user8268
    Commented Sep 14, 2013 at 20:34
  • $\begingroup$ @user8268 Thanks. Now I need to track down the error in my proof. $\endgroup$ Commented Sep 14, 2013 at 20:36

3 Answers 3

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The problem with your proof as written is that $n$ is arbitrary, so when you choose $N$ you are not choosing a fixed value - you need a different $N$ for each $n$.

You could approach this by noting that there is an $N$ with $|a_m|\lt 1$ for $m\gt N$, and therefore $|a_m|^2\lt |a_m|$, which sets up a comparison using the absolute convergence.

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Here's an alternative:

Suppose $\sum \limits_{n=1}^{+\infty}\left(\left|a_n\right|\right)$ converges.

It follows that $(|a_n|)_{n\in \Bbb N}\longrightarrow 0$, therefore there exists $k>0$ such that $\forall n\in \Bbb N(|a_n|<k)$.

Thus $\forall n\in \Bbb N\left(|a_n|^2<k|a_n|\right)$.

Since $\sum \limits_{n=1}^{+\infty}\left(k|a_n|\right)$ converges and due to all the sequences involved being of non-negative real numbers, it follows that $\sum \limits _{n=1}^{+\infty}\left(|a_n|^2\right)$ converges and the result follows.

Note that $\left(a_n\right)_{n\in \Bbb N}$ can be a complex sequence too.

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    $\begingroup$ "$\forall n \in \mathbb{N} \quad |a_n|<k$": I would say something like: $(\exists n_0 \in \mathbb{N})(\forall n \geq n_0) |a_{n}|<k$. It's obvious what you meant, but... $\endgroup$
    – Cortizol
    Commented Sep 14, 2013 at 21:43
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    $\begingroup$ @Cortizol What I said is true. What you said is true. Both are sufficient to make the reasoning work. I opted for the former. $\endgroup$
    – Git Gud
    Commented Sep 14, 2013 at 21:44
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Apply limit test, observe ${a_n^2\over |a_n|}\to0$. Now $\sum |a_n|$ converges $\implies\sum a_n^2$ converges.

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    $\begingroup$ Assuming $a_n$ to be non zero $\endgroup$
    – Vidit
    Commented Jul 30, 2020 at 17:41

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